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\begin{document}
\title{ \Huge{\it Stochastic Calculus for Finance I: The Binomial Asset Pricing Model} \\ {Solution of Exercise Problems} }
\author{Yan Zeng}
\date{Version 1.1, last revised on 2014-10-26}

\maketitle

\begin{abstract}
This is a solution manual for Shreve \cite{Shreve04a}. If you find any typos/errors or have any comments, please email me at zypublic@hotmail.edu.
\end{abstract}

\tableofcontents

\newpage

\section{The Binomial No-Arbitrage Pricing Model}

$\bigstar$ {\bf Comments}:

\medskip

1) Example 1.1.1 illustrates the essence of arbitrage: {\it buy low, sell high}. Since concrete numbers often obscure the nature of things, we review Example 1.1.1 in abstract symbols.

\smallskip

First, the possibility of replicating the payoff of a call option, $(S_1-K)^+$, and its reverse, $-(S_1-K)^+$.

To replicate the payoff $(S_1 - K)^+$ of a call option at time 1, we at time 0 construct a portfolio $(X_0 - \Delta_0 S_0, \Delta_0 S_0)$. At the operational level, we borrow $X_0$, buy $\Delta_0$ shares of stock, and invest the residual amount $(X_0 - \Delta_0 S_0)$ into money market account. The result is a net cash flow of $X_0$ into the portfolio. The replication requirement at time 1 is
\[
(1+r)(X_0 - \Delta_0 S_0) + \Delta_0 S_1 = (S_1 - K)^+.
\]
Plug in $S_1(H)=uS_0$ and $S_1(T)=dS_0$, we obtain a system of two linear equations for two unknowns ($X_0$ and $\Delta_0$) and it has a unique solution as long as $u\ne d$. This is how we obtain the magic number $X_0=1.20$ and $\Delta_0=\frac{1}{2}$ in Example 1.1.1.

To replicate the reverse payoff $-(S_1 − K)^+$ of a call option at time 1, we at time 0 construct a portfolio $(-X_0 + \Delta_0 S_0, -\Delta_0 S_0)$. At the operational level, we short sell $\Delta_0$ shares of stock, invest the income $\Delta_0 S_0$ into money market account, and withdraw a cash amount of $X_0$ from the money market account.
The result is a net cash flow of $X_0$ out of the portfolio.

\smallskip

Second, the realization of arbitrage opportunity through {\it buy low, sell high}.

If the market price $C_0$ of the call option is greater than $X_0$, we just sell the call option for $C_0$ (sell high), spend $X_0$ on constructing the synthetic call option (buy low), and take the residual amount $(C_0-X_0)$ as arbitrage profit. At time 1, the payoff of short position in the call option will cancel out with the payoff of the synthetic call option.

If the market price $C_0$ of the call option is less than $X_0$, we buy the call option for $C_0$ (buy low) and set up the portfolio $(-X_0 + \Delta_0 S_0, -\Delta_0 S_0)$ at time 0, which allows us to withdraw a cash amount of $X_0$ (sell high). The net cash flow $(X_0-C_0)$ at time 0 is taken as arbitrage profit. At time 1, the payoff of long position in the call option will cancel out with the payoff of the portfolio $(-X_0 + \Delta_0 S_1, -\Delta_0 S_1)$.

\medskip

2) The essence of Definition 1.2.3 is that we can find a replicating portfolio $(\Delta_0$, $\cdots$, $\Delta_{N-1})$ and ``define" the portfolio's value at time $n$ as the price of the derivative security at time $n$. The rationale is that if $P(\omega_1\omega_2\cdots\omega_N)>0$ and the price of the derivative security at time $n$ does not agree with the portfolio's value, we can make an arbitrage on sample path $\omega_1\omega_2\cdots\omega_N$, starting from time $n$. For a formal presentation of this argument, we refer to Delbaen and Schachermayer \cite{DS06}, Chapter 2.

Also note the definition needs the uniqueness of the replicating portfolio. This is guaranteed in the binomial model as seen from the uniqueness of solution of equation (1.1.3)-(1.1.4).

Finally, we note the wealth equation (1.2.14) can be written as
\[
\frac{X_{n+1}}{(1+r)^{n+1}} = \frac{X_n}{(1+r)^n} + \Delta_n \left[\frac{S_{n+1}}{(1+r)^{n+1}} - \frac{S_n}{(1+r)^n}\right]
\]
This leads to a representation by discrete stochastic integral:
\[
\widetilde X_T = X_0 + (\Delta\cdot \widetilde S)_T,
\]
where $\widetilde X_n = \frac{X_n}{(1+r)^n}$ and $\widetilde S_n = \frac{S_n}{(1+r)^n}$, $n=1, 2, \cdots, N$.

\bigskip

\noindent $\blacktriangleright$  {\bf Exercise 1.1.} Assume the one-period binomail market of Section 1.1 that both $H$ and $T$ have positive probability of occurring. Show that condition (1.1.2) precludes arbitrage. In other words, show that if $X_0=0$ and
\[
X_1 = \Delta_0 S_1 + (1+r)(X_0 - \Delta_0 S_0),
\]
then we cannot have $X_1$ strictly positive with positive probability unless $X_1$ is strictly negative with positive probability as well, and this is the case regardless of the choice of the number $\Delta_0$.

\begin{proof}
Note the random stock price has only two states at time 1, $H$ and $T$. So ``we cannot have $X_1$ strictly positive with positive probability unless $X_1$ is strictly negative with positive probability as well" can be succinctly summarized as
\[
``X_1(H) > 0 \Rightarrow X_1(T) < 0 \mbox{ and } X_1(T) > 0 \Rightarrow X_1(H) < 0".
\]

We prove a slightly more general version of the problem to expose the nature of no-arbitrage:
\[
``X_1(H) > (1+r)X_0 \Rightarrow X_1(T) < (1+r)X_0 \mbox{ and } X_1(T) > (1+r)X_0 \Rightarrow X_1(H) < (1+r)X_0".
\]
Note this is indeed a generalization since the original problem assumes $X_0=0$.

For a formal proof, we write $X_1$ as
\[
X_1 = \Delta_0S_0 \left(\frac{S_1 - S_0}{S_0} - r\right) + (1+r)X_0.
\]
Then
\[
X_1(H) - (1+r)X_0 = \Delta_0 S_0 [u-(1+r)]
\]
and
\[
X_1(T) - (1+r)X_0 = \Delta_0 S_0 [d-(1+r)].
\]

Given the condition $d<1+r<u$, the factor $S_0 [u-(1+r)]$ is positive and the factor $S_0 [d-(1+r)]$ is negative. So
\[
X_1(H) > (1+r)X_0 \Rightarrow \Delta_0 > 0 \Rightarrow X_1(T) < (1+r) X_0
\]
and
\[
X_1(T) > (1+r)X_0 \Rightarrow \Delta_0 < 0 \Rightarrow X_1(H) < (1+r) X_0.
\]
This concludes our proof.
\end{proof}

\begin{remark}
The textbook (page 2-3) has shown ``negation of $d < 1+ r < u$ $\Rightarrow$ arbitrage", or equivalently,
``no arbitrage $\Rightarrow$ $d < 1+ r < u$". This exercise problem asks us to prove ``$d < 1+ r < u$ $\Rightarrow$ no arbitrage".
\end{remark}

\begin{remark}
In the equation $X_1 = \Delta_0 S_1 + (1+r)(X_0 - \Delta_0 S_0)$, the first term $\Delta_0 S_1$ is the value of the stock position at time 1 while the second term $(1+r)(X_0 - \Delta_0 S_0)$ is the value of the money market account at time 1.
\end{remark}

\begin{remark}
The condition $X_0=0$ in the original problem formulation is not really essential, as far as
a proper definition of arbitrage can be given.
Indeed, for the one-period binomial model, we can define arbitrage
as a trading strategy such that $P(X_1\ge X_0(1+r))=1$ and $P(X_1>
X_0(1+r))>0$. That is, arbitrage is a trading strategy whose return beats the risk-free rate. See Shreve \cite[page~254]{Shreve04b} Exercise 5.7 for this more general definition.
\end{remark}

\begin{remark} Note the condition $d < 1+r < u$ is just $\frac{S_1(T)-S_0}{S_0} < r < \frac{S_1(H)-S_0}{S_0}$: the return of the stock investment is not guaranteed to be greater or less than the risk-free rate. This agrees with the intuition that ``arbitrage is a trading strategy that is guaranteed to beat the risk-free investment".
\end{remark}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 1.2.} Suppose in the situation of Example 1.1.1 that the option sells for 1.20 at time zero. Consider an agent who begins with wealth $X_0=0$ and at time zero buys $\Delta_0$ shares of stock and $\Gamma_0$ options. The numbers $\Delta_0$ and $\Gamma_0$ can be either positive or negative or zero. This leaves the agent with a cash position of $-4\Delta_0-1.20\Gamma_0$. If this is positive, it is invested in the money market; if it is negative, it represents money borrowed from the money market. At time one, the value of the agent's portfolio of stock, option, and money market assets is
\[
X_1 = \Delta_0 S_1 + \Gamma_0 (S_1-5)^+ - \frac{5}{4}(4\Delta_0+1.20\Gamma_0).
\]
Assume that both $H$ and $T$ have positive probability of occurring. Show that if there is a positive probability that $X_1$ is positive, then there is a positive probability that $X_1$ is negative. In other words, one cannot find an arbitrage when the time-zero price of the option is 1.20.

 \begin{proof}$X_1(u)=\Delta_0\times 8+\Gamma_0\times
3-\frac{5}{4}(4\Delta_0+1.20\Gamma_0)=3\Delta_0+1.5\Gamma_0$, and
$X_1(d)=\Delta_0\times
2-\frac{5}{4}(4\Delta_0+1.20\Gamma_0)=-3\Delta_0-1.5\Gamma_0$. That
is, $X_1(u)=-X_1(d)$. So if there is a positive probability that
$X_1$ is positive, then there is  a positive probability that $X_1$
is negative. This finishes the proof of the original problem.

\smallskip

Since the use of numbers often obscures the nature of a problem, and since the notation of this exercise problem is rather confusing (given the notation in Example 1.1.1), we shall re-state the problem in different notation and rewrite the proof in abstract symbols.

First, a re-statement of the problem in different notation: ``If the option is priced at the replication cost $C_0$ (which is denoted by $X_0$ in Example 1.1.1), we cannot find arbitrage by investing in tradable securities (stock, option, and money market account). Formally, suppose an investor borrows money to buy $\alpha$ shares of stock and $\beta$ options at time 0, the portfolio thus constructed is $(-\alpha S_0 - \beta C_0,\alpha S_0 + \beta C_0)$. Its value at time 1 becomes
\[
X_1 = \alpha S_1 + \beta V_1 - (1+r)(\alpha S_0 + \beta C_0)
\]
where $V_1$ is the option payoff $(S_1 - K)^+$. Prove $P(X_1 > 0) >0 \Rightarrow P(X_1 < 0) > 0$."

Second, our proof in abstract symbols. Recall for the replication cost $C_0$, there exists some $\delta >0$ such that
\[
 (1+r)(C_0 - \delta S_0) + \delta S_1 \equiv V_1.
\]
Plug this into the expression of $X_1$, we have
\begin{eqnarray*}
X_1
&=& \alpha S_1 + \beta V_1 - (1+r)(\alpha S_0 + \beta C_0) \\
&=& \alpha S_1 + \beta [(1+r)(C_0 - \delta S_0) + \delta S_1] - (1+r)(\alpha S_0 + \beta C_0) \\
&=& \alpha S_1 + \beta(1+r)C_0 - \beta \delta S_0 (1+r) + \delta \beta S_1 - (1+r) \alpha S_0 - (1+r)\beta C_0 \\
&=& (\alpha + \delta \beta)S_1 - S_0(1+r)(\alpha + \beta\delta) \\
&=& (\alpha + \beta\delta)S_0 \left[\frac{S_1}{S_0} - (1+r)\right].
\end{eqnarray*}
Since $S_1/S_0 = u$ or $d$ and $d < 1+r < u$, we conclude $X_1(H)$ and $X_1(T)$ have opposite signs. This concludes our proof.
\end{proof}

\begin{remark}
Example 1.1.1 has shown that ``no arbitrage $\Rightarrow$ the time-zero price of the option is 1.20" by proving ``the time-zero price of the option is not 1.20 $\Rightarrow$ there exists arbitrage via linear combination of investments in stock, option, and money market account".

This exercise problem asks us to prove ``the time-zero price of the option is 1.20 $\Rightarrow$ there exists no arbitrage via the linear combination of investments in stock, option and money market account". Although logically, ``linear combination of tradable securities" is only a special way of constructing portfolios, in practice it is the only way. So it's all right to use this special form of arbitrage to stand for general arbitrage.
\end{remark}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 1.3.} In the one-period binomial model of Section 1.1, suppose we want to determine the price at time zero of the derivative security $V_1 = S_1$ (i.e., the derivative security pays off the stock price.) (This can be regarded as a European call with strike price $K=0$). What is the time-zero price $V_0$ given by the risk-neutral pricing formula (1.1.10)?

\begin{solution}
$V_0=\frac{1}{1+r}\left[\frac{1+r-d}{u-d}S_1(H)+\frac{u-1-r}{u-d}S_1(T)\right]=\frac{S_0}{1+r}\left(\frac{1+r-d}{u-d}u+\frac{u-1-r}{u-d}d\right)=S_0$.
This is not surprising, since this is exactly the cost of
replicating $S_1$, via the buy-and-hold strategy.
\end{solution}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 1.4.} In the proof of Theorem 1.2.2, show under the induction hypothesis that
\[
X_{n+1}(\omega_1 \omega_2 \cdots \omega_n T) = V_{n+1}(\omega_1 \omega_2 \cdots \omega_n T).
\]

\begin{proof}
\begin{eqnarray*}
X_{n+1}(T)&=&\Delta_ndS_n+(1+r)(X_n-\Delta_nS_n)\\
&=&\Delta_nS_n(d-1-r)+(1+r)V_n\\
&=&\frac{V_{n+1}(H)-V_{n+1}(T)}{u-d}(d-1-r)+(1+r)\frac{\tilde
pV_{n+1}(H)+\tilde qV_{n+1}(T)}{1+r}\\
&=&\tilde p [V_{n+1}(T)-V_{n+1}(H)] +\tilde pV_{n+1}(H)+\tilde
qV_{n+1}(T)\\
&=&\tilde pV_{n+1}(T)+\tilde qV_{n+1}(T)\\
&=&V_{n+1}(T).
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 1.5.} In Example 1.2.4, we considered an agent who sold the look-back option for $V_0 = 1.376$ and bought $\Delta_0 = 0.1733$ shares of stock at time zero. At time one, if the stock goes up, she has a portfolio valued at $V_1(H) = 2.24$. Assume that she now takes a position of $\Delta_1(H) = \frac{V_2(HH)-V_2(HT)}{S_2(HH)-S_2(HT)}$ in the stock. Show that, at time two, if the stock goes up again, she will have a portfolio valued at $V_2(HH)=3.20$, whereas if the stock goes down, her portfolio will be worth $V_2(HT) = 2.40$. Finally, under the assumption that the stock goes up in the first period and down in the second period, assume the agent takes a position of $\Delta_2(HT) = \frac{V_3(HTH)-V_3(HTT)}{S_3(HTH)-S_3(HTT)}$ in the stock. Show that, at time three, if the stock goes up in the third period, she will have a portfolio valued at $V_3(HTH)=0$, whereas if the stock goes down, her portfolio will be worth $V_3(HTT)=6$. In other words, she has hedged her short position in the option.

\begin{proof}
First, on the path $\omega_1 = H$, the investor's portfolio is worth of
\begin{eqnarray*}
X_1(H) = (1+r)(X_0 - \Delta_0 S_0) + \Delta_0 S_1(H) = (1+0.25)(1.376-0.1733\cdot 4) + 0.1733 \cdot 8 = 2.24 = V_1(H).
\end{eqnarray*}
If on the path $\omega_1=H$, the investor takes a position of
\[
\Delta_1(H) = \frac{V_2(HH)-V_2(HT)}{S_2(HH)-S_2(HT)} = \frac{3.20 - 2.40}{16 - 4} = 0.0667
\]
in the stock, then on the path $\omega_1\omega_2=HH$, the investor's portfolio is worth of
\begin{eqnarray*}
X_2(HH) &=& (1+r)[X_1(H) - \Delta_1(H)S_1(H)] + \Delta_1(H)S_2(HH) \\
&=& (1+0.25)(2.24 - 0.0667 \cdot 8) + 0.0667 \cdot 16 \\
&=& 3.2 \\
&=& V_2(HH),
\end{eqnarray*}
while on the path $\omega_1\omega_2=HT$, the investor's portfolio is worth of
\begin{eqnarray*}
X_2(HT) &=& (1+r)[X_1(H) - \Delta_1(H)S_1(H)] + \Delta_1(H)S_2(HT) \\
&=& (1+0.25)(2.24 - 0.0667 \cdot 8) + 0.0667 \cdot 4 \\
&=& 2.4 \\
&=& V_2(HT).
\end{eqnarray*}

Second, if on the path $\omega_1\omega_2=HT$, the agent takes a position
\[
\Delta_2(HT) = \frac{V_3(HTH)-V_3(HTT)}{S_3(HTH)-S_3(HTT)} = \frac{0-6}{8-2}=-1
\]
in the stock, then on the path $\omega_1\omega_2\omega_3=HTH$, the investor's portfolio is worth of
\begin{eqnarray*}
X_3(HTH) &=& (1+r) [X_2(HT)-\Delta_2(HT)S_2(HT)] + \Delta_2(HT) S_3(HTH) \\
&=& (1+0.25)[2.4 - (-1)\cdot 4] + (-1) \cdot 8 \\
&=& 0 \\
&=& V_3(HTH),
\end{eqnarray*}
while on the path $\omega_1\omega_2\omega_3=HTT$, the investor's portfolio is worth of
\begin{eqnarray*}
X_3(HTT) &=& (1+r) [X_2(HT)-\Delta_2(HT)S_2(HT)] + \Delta_2(HT) S_3(HTT) \\
&=& (1+0.25)[2.4 - (-1)\cdot 4] + (-1) \cdot 2 \\
&=& 6 \\
&=& V_3(HTT).
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 1.6.} {\bf (Hedging a long position-one period).} Consider a bank that has a long position in the European call written on the stock price in Figure 1.1.2. The call expires at time one and has strike price $K=5$. In Section 1.1, we determined the time-zero price of this call to be $V_0 = 1.20$. At time zero, the bank owns this option, while ties up capital $V_0 = 1.20$. The bank wants to earn the interest rate $25\%$ on this capital until time one (i.e., without investing any more money, and regardless of how the coin tossing turns out, the bank wants to have
\[
\frac{5}{4} \cdot 1.20 = 1.50
\]
at time one, after collecting the payoff from the option (if any) at time one). Specify how the bank's trader should invest in the stock and money markets to accomplish this.

\begin{solution} The bank's trader should set up a
replicating portfolio whose payoff is the opposite of the option's
payoff. More precisely, we solve the equation
\[
(1+r)(X_0-\Delta_0S_0)+\Delta_0S_1=-(S_1-K)^+.
\]Then $X_0=-1.20$ and $\Delta_0=-\frac{1}{2}$ since this equation is a linear equation of $X_0$ and $\Delta_0$. The solution means the
trader should sell short 0.5 share of stock, put the income 2 into a
money market account, and then transfer 1.20 into a separate money
market account. At time one, the portfolio consisting of a short
position in stock and $0.8(1+r)$ in money market account will cancel
out with the option's payoff. In the end, we end up with $1.20(1+r)$
in the separate money market account.
\end{solution}

\begin{remark} This problem illustrates why we are interested in
hedging a long position. In case the stock price goes down at time
one, the option will expire worthless. The initial amount of money
1.20 paid at time zero will be wasted. By hedging, we convert the
option back into liquid assets (cash and stock) which guarantees a
sure payoff at time one. As to why we
hedge a short position (as a writer), see Wilmott \cite[page~11-13]{Wilmott95},
1.4 What are Options For?
\end{remark}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.7.} {\bf (Hedging a long position-multiple periods).} Consider a bank that has a long position in the lookback option of Example 1.2.4. The bank intends to hold this option until expiration and receive the payoff $V_3$. At time zero, the bank has capital $V_0 = 1.376$ tied up in the option and wants to earn the interest rate of $25\%$ on this capital until time three (i.e., without investing any more money, and regardless of how the coin tossing turns out, the bank wants to have
\[
\left(\frac{5}{4}\right)^3 \cdot 1.376 = 2.6875
\]
at time three, after collecting the payoff from the lookback option at time three). Specify how the bank's trader should invest in the stock and the money market account to accomplish this.

\begin{solution} The idea is the same as Exercise 1.6. The
bank's trader only needs to set up the reverse of the replicating
trading strategy described in Example 1.2.4. More precisely, he
should short sell 0.1733 share of stock, invest the income 0.6933
into money market account, and transfer 1.376 into a separate money
market account. The portfolio consisting a short position in stock
and 0.6933-1.376 in money market account will replicate the opposite
of the option's payoff. After they cancel out, we end up with
$1.376(1+r)^3$ in the separate money market account.
\end{solution}

\medskip

\pagebreak

\noindent $\blacktriangleright$ {\bf Exercise 1.8.} {\bf (Asian option).} Consider the three-period model of Example 1.3.1,\footnote{The textbook said ``Example 1.2.1" by mistake.} with $S_0 = 4$, $u=2$, $d=\frac{1}{2}$, and take the interest rate $r=\frac{1}{4}$, so that $\tilde p = \tilde q = \frac{1}{2}$. For $n=0, 1, 2, 3$, define $Y_n = \sum_{k=0}^n S_k$ to be the sum of the stock prices between times zero and $n$. Consider an {\it Asian call option} that expires at time three and has strike $K=4$ (i.e., whose payoff at time three is $\left(\frac{1}{4}Y_3-4\right)^+$). This is like a European call, except the payoff of the option is based on the average stock price rather than the final stock price. Let $v_n(s,y)$ denote the price of this option at time $n$ if $S_n=s$ and $Y_n=y$. In particular, $v_3(s,y) = \left(\frac{1}{4}y-4\right)^+$.

\smallskip

\noindent (i) Develop an algorithm for computing $v_n$ recursively. In particular, write a formula for $v_n$ in terms of $v_{n+1}$.

\begin{solution} By risk-neutral pricing formula,
\[
v_n(s,y)=\frac{1}{1+r} \left[\tilde p v_{n+1}(us,y+us)+ \tilde q v_{n+1}\left(ds,y+ds\right) \right] = \frac{2}{5} \left[ v_{n+1}(2s,y+2s)+v_{n+1}\left(\frac{s}{2},y+\frac{s}{2}\right) \right].
\]
\end{solution}

\noindent (ii) Apply the algorithm developed in (i) to compute $v_0(4,4)$, the price of the Asian option at time zero.

\begin{solution}

\begin{figure}[h] % Place the float at the same point it occurs in the source text (however, not exactly at the spot)
\centering
% Set the overall layout of the tree
\tikzstyle{level 1}=[level distance=3.5cm, sibling distance=7cm, ->]
\tikzstyle{level 2}=[level distance=3.5cm, sibling distance=3.5cm, ->]
\tikzstyle{level 3}=[level distance=3.5cm, sibling distance=1.5cm, ->]

% Define styles for bags and leafs
\tikzstyle{bag} = [text width=7em, text centered]
\tikzstyle{end} = [text width=7em, text centered]

\begin{tikzpicture}[grow=right, sloped]
\node[bag]{$S_0=4$ $Y_0=4$ $v_0(4,4)=1.216$}
    child {
        node[bag]{$S_1(T)=2$ $Y_1(T)=6$ $v_1(2,6)=0.08$}
            child {
                node[bag]{$S_2(TT)=1$ $Y_2(TT)=7$ $v_2(1,7)=0$}
                    child {
                        node[end]{$S_3(TTT)=0.5$ $Y_3(TTT)=7.5$ $v_3(0.5,7.5)=0$}
                    }
                    child {
                        node[end]{$S_3(TTH)=2$ $Y_3(TTH)=9$ $v_3(2,9)=0$}
                    }
            }
            child {
                node[bag]{$S_2(TH)=4$ $Y_2(TH)=10$ $v_2(4,10)=0.2$}
                    child {
                        node[end]{$S_3(THT)=2$ $Y_3(THT)=12$ $v_3(2,12)=0$}
                    }
                    child {
                        node[end]{$S_3(THH)=8$ $Y_3(THH)=18$ $v_3(8,18)=0.5$}
                    }
            }
    }
    child {
        node[bag]{$S_1(H)=8$ $Y_1(H)=12$ $v_1(8,12)=2.96$}
            child {
                node[bag]{$S_2(HT)=4$ $Y_2(HT)=16$ $v_2(4,16)=1$}
                    child {
                        node[end]{$S_3(HTT)=2$ $Y_3(HTT)=18$ $v_3(2,18)=0.5$}
                    }
                    child {
                        node[end]{$S_3(HTH)=8$ $Y_3(HTH)=24$ $v_3(8,24)=2$}
                    }
            }
            child {
                node[bag]{$S_2(HH)=16$ $Y_2(HH)=28$ $v_2(16,28)=6.4$}
                    child {
                        node[end]{$S_3(HHT)=8$ $Y_3(HHT)=36$ $v_3(8,36)=5$}
                    }
                    child {
                        node[end]{$S_3(HHH)=32$ $Y_3(HHH)=60$ $v_3(32,60)=11$}
                    }
            }
    }
;
\end{tikzpicture}
\caption{{\bf Exercise 1.8.} Asian option.}
\end{figure}
\end{solution}

\noindent (iii) Provide a formula for $\delta_n(s,y)$, the number of shares of stock that should be held by the replicating portfolio at time $n$ if $S_n = s$ and $Y_n = y$.

\begin{solution}
\[
\delta_n(s,y)=\frac{v_{n+1}(us,y+us)-v_{n+1}(ds,y+ds)}{(u-d)s}.
\]
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 1.9.} {\bf (Stochastic volatility, random interest rate).} Consider a binomial pricing model, but at each time $n\ge 1$, the ``up factor" $u_n(\omega_1 \omega_2 \cdots \omega_n)$, the ``down factor" $d_n(\omega_1 \omega_2 \cdots \omega_n)$, and the interest rate $r_n(\omega_1 \omega_2 \cdots \omega_n)$ are allowed to depend on $n$ and on the first $n$ coin tosses $\omega_1 \omega_2 \cdots \omega_n$. The initial up factor $u_0$, the initial down factor $d_0$, and the initial interest rate $r_0$ are not random. More specifically, the stock price at time one is given by
\[
S_1(\omega_1) = \begin{cases}
u_0 S_0 & \mbox{if $\omega_1 = H$,} \\
d_0 S_0 & \mbox{if $\omega_1 = T$,}
\end{cases}
\]
and, for $n\ge 1$, the stock price at time $n+1$ is given by
\[
S_{n+1}(\omega_1 \omega_2 \cdots \omega_n \omega_{n+1}) = \begin{cases}
u_n(\omega_1\omega_2\cdots\omega_n)S_n(\omega_1\omega_2\cdots\omega_n) & \mbox{if $\omega_{n+1}=H$,} \\
d_n(\omega_1\omega_2\cdots\omega_n)S_n(\omega_1\omega_2\cdots\omega_n) & \mbox{if $\omega_{n+1}=T$.}
\end{cases}
\]
One dollar invested in or borrowed from the money market at time zero grows to an investment or debt of $1+r_0$ at time one, and, for $n \ge 1$, one dollar invested in or borrowed from the money market at time $n$ grows to an investment or debt of $1+r_n(\omega_1\omega_2\cdots\omega_n)$ at time $n+1$. We assume that for each $n$ and for all $\omega_1\omega_2\cdots\omega_n$, the no-arbitrage condition
\[
0 < d_n(\omega_1\omega_2\cdots\omega_n) < 1 + r_n(\omega_1\omega_2\cdots\omega_n) < u_n(\omega_1\omega_2\cdots\omega_n)
\]
holds. We also assume that $0 < d_0 < 1+ r_0 < u_0$.

\smallskip

\noindent (i) Let $N$ be a positive integer. In the model just described, provide an algorithm for determining the price at time zero for a derivative security that at time $N$ pays off a random amount $V_N$ depending on the result of the first $N$ coin tosses.

\begin{solution} Similar to Theorem 1.2.2, but replace
$r$, $u$ and $d$ everywhere with $r_n$, $u_n$ and $d_n$. More
precisely, set $\tilde p_n=\frac{1+r_n-d_n}{u_n-d_n}$ and
$\tilde q_n=1-\widetilde p_n$. Then
\[
V_n=\frac{\tilde p_nV_{n+1}(H)+\tilde q_nV_{n+1}(T)}{1+r_n}.
\]
\end{solution}

\noindent (ii) Provide a formula for the number of shares of stock that should be held at each time $n$ ($0 \le n \le N-1$) by a portfolio that replicates the derivatives security $V_N$.

\begin{solution}
$\Delta_n=\frac{V_{n+1}(H)-V_{n+1}(T)}{S_{n+1}(H)-S_{n+1}(T)}=\frac{V_{n+1}(H)-V_{n+1}(T)}{(u_n-d_n)S_n}$.
\end{solution}

\noindent (iii) Suppose the initial stock price is $S_0 = 80$, with each head the stock price increases by 10, and with each tail the stock price decreases by 10. In other words, $S_1(H) = 90$, $S_1(T) = 70$, $S_2(HH)=100$, etc. Assume the interest rate is always zero. Consider a European call with strike price 80, expiring at time five. What is the price of this call at time zero?

\begin{solution}
$u_n=\frac{S_{n+1}(H)}{S_n}=\frac{S_n+10}{S_n}=1+\frac{10}{S_n}$ and
$d_n=\frac{S_{n+1}(T)}{S_n}=\frac{S_n-10}{S_n}=1-\frac{10}{S_n}$. So
the risk-neutral probabilities at time $n$ are $\tilde
p_n=\frac{1-d_n}{u_n-d_n}=\frac{1}{2}$ and $\tilde q_n=\frac{1}{2}$.

To price the European call, we only need to focus on the nodes at time 5 of the binomial tree since $\tilde p_n$ and $\tilde q_n$ are constants. In order to have non-zero payoffs, a node at time 5 must have at least 3 $H$'s. For such a node, we have

\begin{center}
    \begin{tabular}{|l|l|l|}
    \hline
    number of $H$'s & option payoff              & risk-neutral probability of paths\\
    \hline
    $3$             & $(80 + 30 - 20) - 80 = 10$ &  $\frac{1}{2^5} \cdot \frac{5!}{3!2!} = \frac{10}{32} $\\
    \hline
    $4$             & $(80 + 40 - 10) - 80 = 30$ &  $\frac{1}{2^5} \cdot 5 = \frac{5}{32}$ \\
    \hline
    $5$             & $(80 + 50) - 80 = 50$      & $\frac{1}{2^5} = \frac{1}{32}$ \\
    \hline
    \end{tabular}
\end{center}

So the price of the call at time zero is
\[
\frac{1}{32} \cdot 50 + \frac{5}{32} \cdot 30 + \frac{10}{32} \cdot 10 = \frac{300}{32} = 9.375.
\]
\end{solution}


\section{Probability Theory on Coin Toss Space}

$\bigstar$ {\bf Comments}:

\medskip

1) The second proof of Theorem 2.4.4 also works for the random interest rate model of Exercise 1.9, as far as $\frac{S_{n+1}}{(1+r)^{n+1}}$ is replaced by $\frac{S_{n+1}}{(1+r_0)\cdots(1+r_n)}$. The requirement on the risk-neutral probability $\widetilde {\mathbb P}$ is that
\[
\widetilde {\mathbb P}(w_{n+1}=H|\omega_1,\cdots, \omega_n) := \tilde p_n = \frac{1+r_n-d_n}{u_n-d_n}
\]
and
\[
\widetilde {\mathbb P}(w_{n+1}=T|\omega_1,\cdots, \omega_n) := 1-\tilde p_n = \tilde q_n.
\]
Results in measure theory guarantee the existence and uniqueness of such a probability measure $\widetilde {\mathbb P}$ on the sample space $\Omega=\{w: (w_1, \cdots, w_N)\}$ for a given family of conditional probabilities $(\tilde p_n, \tilde q_n)_{n=1}^N$.\footnote{See, for example, Shiryaev \cite[page~249]{Shiryaev95}, Theorem 2 (Ionescu Tulcea's Theorem on Extending a Measure and the Existence of a Random Sequence).\label{footnote:Ionescu-Tulcea}} With this setup, we have
\[
\widetilde {\mathbb E}_n\left[\frac{S_{n+1}}{S_n}\right] = u_n\tilde p_n + d_n \tilde q_n = (1+r_n).
\]
In other words, the stock price process, when discounted by the random interest rates, is a martingale under $\widetilde {\mathbb P}$. When $r_n$'s are deterministic, we are back to the case where $\omega_n$'s are independent of each other. This unifies the deterministic and random interest rate models.

\medskip

2) On Theorem 2.4.8 (Cash flow valuation): by Theorem 2.4.7, it is natural to ``conjecture" that the no-arbitrage price process of the derivative security is given by the risk-neutral pricing formula (2.4.13). However, pricing always needs to be justified by hedging/replication. We need to find a portfolio process which replicates the cash flows while equals to $(V_n)_{n=0}^N$ in value. The key insight to the construction of such a replicating portfolio is the wealth equation (2.4.16)
\[
X_{n+1} = \Delta_n S_{n+1} + (1+r) (X_n - C_n - \Delta_nS_n),
\]
which clearly indicates that the wealth process $(X_n)_{n=1}^N$ produces a cash outflow of $C_n$ at step $n$.

Note the presentation of Theorem 2.4.8 follows the following flow of logic:
\[
\mbox{define $V_n$'s}
\Rightarrow \mbox{define $\Delta_n$'s}
\Rightarrow  \mbox{define $X_n$'s}
\Rightarrow \mbox{prove $X_n=V_n$}.
\]
We could have taken a different, yet more natural path of logic, namely
\begin{eqnarray*}
&           & \mbox{define $X_n$'s and $\Delta_n$'s simultaneously} \\
&\Rightarrow& \mbox{prove formula (2.4.14) holds with $V_n$'s replaced by $X_n$'s} \\
&\Rightarrow& \mbox{prove formula (2.4.13) holds with $V_n$'s replaced by $X_n$'s} \\
&\Rightarrow& \mbox{define $V_n=X_n$}.
\end{eqnarray*}
Indeed, we define $X_N=C_N$ and solve for $X_{N-1}$ and $\Delta_{N-1}$ in the following equation
\[
X_N = \Delta_{N-1}S_N + (1+r)(X_{N-1} - C_{N-1} - \Delta_{N-1}S_{N-1}).
\]
By imitating the trick of (1.1.3)-(1.1.8) (page 6), we obtain
\[
\Delta_{N-1}(\omega_1\cdots \omega_{N-1})=\frac{C_N(\omega_1\cdots \omega_{N-1} H)-C_N(\omega_1\cdots \omega_{N-1}T)}{S_N(\omega_1\cdots \omega_{N-1} H) - S_N(\omega_1\cdots \omega_{N-1} T)}
\]
\begin{eqnarray*}
X_{N-1} &=& C_{N-1} + \frac{1}{1+r}\left[\tilde p X_N(\omega_1\cdots \omega_{N-1} H) + \tilde q X_N(\omega_1\cdots \omega_{N-1} T)\right] \\
&=& C_{N-1} + \widetilde {\mathbb E}_n \left[\frac{X_N}{1+r}\right]\\
&=& \widetilde {\mathbb E}_n\left[\sum_{k=N-1}^N\frac{C_k}{(1+r)^{k-(N-1)}}\right].
\end{eqnarray*}
Working backward by induction, we can provide definitions for each $X_n$ and $\Delta_n$, and prove for each $n$
\[
X_n = C_n + \widetilde {\mathbb E}_n\left[\frac{X_{n+1}}{1+r}\right] = \widetilde {\mathbb E}_n \left[\sum_{k=n}^N\frac{C_k}{(1+r)^{k-n}}\right].
\]

Finally, we comment that Theorem 1.2.2 is a special case of Theorem 2.4.8, with $C_0=C_1=\cdots=C_{N-1}=0$.

\medskip

3) The textbook gives readers the impression that Markov property is preserved under the risk-neutral probability, at least in the setting of binomial model. A general result in this regard is recently announced in Schmock \cite{Schmock13}:
\begin{theorem}[Uwe Schmock and Ismail Cetin G\"{u}l\"{u}m] Let $(\Omega, {\cal F}, \{{\cal F}_t\}_{t=0,\cdots,T}, {\mathbb P})$ be a (general) filtered probability space and let $S = \{S_t\}_{t\in \{0,\cdots, T\}}$ be an adapted, ${\mathbb R}^d$-valued discounted asset price process, which has the $k$-multiple Markov property w.r.t. $\mathbb P$. Then the following properties are equivalent:

\noindent (a) The financial market model is free of arbitrage.

\noindent (b) There exists a probability measure ${\mathbb P}^*$ on $(\Omega, {\cal F}, {\mathbb P})$ such that

$\bullet$ ${\mathbb P}^* \sim {\mathbb P}$ with $\varrho:=d{\mathbb P}^*/d{\mathbb P} \in L^{\infty}(\Omega, {\cal F}_T, {\mathbb P})$.

$\bullet$ {\it Integrability}: $S_t \in L^1(\Omega, {\cal F}_t, {\mathbb P}^*)$ for all $t\in \{0,\cdots, T\}$.

$\bullet$ Martingale property w.r.t. ${\mathbb P}^*$:
\[
{\mathbb E}_{{\mathbb P}^*}[S_t | {\cal F}_{t-1}] \overset{a.s.}{=} S_{t-1} \;\; \mbox{for all $t \in \{1, \cdots, T\}$}.
\]

$\bullet$ $k$-multiple Markov property: For all $B \in {\cal E}$ and $t\in \{k, \cdots, T\}$
\[
{\mathbb P}^*(S_t \in B | {\cal F}_{t-1}) \overset{a.s.}{=} {\mathbb P}^*(S_t \in B | S_{t-1}, S_{t-2}, \cdots, S_{t-k}).
\]
\end{theorem}

For continuous time financial models, it is well-known that for Lipschitz continuous $f$, $g$, the stochastic differential equation of K. It\^{o}
\[
X_t = X_0 + \int_0^t f(s,X_s)dW_s + \int_0^t g(s,X_s)ds
\]
has a unique solution which is a Markov process with continuous paths. Moreover if $f$ and $g$ satisfy $f(t,x)=f(x)$, $g(t,x)=g(x)$, then $X$ is a time homogenous strong Markov process. These results combined with Girsanov's Theorem (Shreve \cite{Shreve04b} Chapter 5) will preserve Markov property of discounted asset process under the risk-neutral measure.

\bigskip

\noindent $\blacktriangleright$  {\bf Exercise 2.1.} Using Definition 2.1.1, show the following.

\smallskip

\noindent (i) If $A$ is an event and $A^c$ denotes its complement, then $\mathbb P(A^c) = 1- \mathbb P(A)$.

\begin{proof}$\mathbb P(A^c)+\mathbb P(A)=\sum_{\omega\in A^c}\mathbb P(\omega)+\sum_{\omega\in A}\mathbb P(\omega)=\sum_{\omega\in \Omega}\mathbb P(\omega)=1$.
\end{proof}

\noindent (ii) If $A_1$, $A_2$, $\cdots$, $A_N$ is a finite set of events, then
\begin{equation}
\mathbb P(\cup_{n=1}^N A_n) \le \sum_{n=1}^N \mathbb P(A_n).\tag{2.8.1}
\end{equation}
If the events $A_1$, $A_2$, $\cdots$, $A_N$ are disjoint, then equality holds in (2.8.1).

\begin{proof}By induction, it suffices to work on the case
$N=2$. When $A_1$ and $A_2$ are disjoint, $\mathbb P(A_1\cup
A_2)=\sum_{\omega\in A_1\cup A_2}\mathbb P(\omega)=\sum_{\omega\in
A_1}\mathbb P(\omega)+\sum_{\omega\in A_2}\mathbb P(\omega)=\mathbb P(A_1)+\mathbb P(A_2)$. When
$A_1$ and $A_2$ are arbitrary, using the result when they are
disjoint, we have $\mathbb P(A_1\cup A_2)=\mathbb P((A_1-A_2)\cup
A_2)=\mathbb P(A_1-A_2)+\mathbb P(A_2)\le P(A_1)+\mathbb P(A_2)$.
\end{proof}

\medskip

\noindent $\blacktriangleright$  {\bf Exercise 2.2.} Consider the stock price $S_3$ in Figure 2.3.1.

\smallskip

\noindent (i) What is the distribution of $S_3$ under the risk-neutral probabilities $\tilde p = \frac{1}{2}$, $\tilde q = \frac{1}{2}$.

\begin{solution} Under the risk-neutral probability, the distribution of $\omega_{n+1}$ conditioning on $\omega_1$, $\cdots$, $\omega_n$ is deterministic
\[
\widetilde {\mathbb P}(\omega_{n+1}=H|\omega_1\omega_2\cdots\omega_n) := \tilde p = \frac{1+r-d}{u-d}, \;\;
\widetilde {\mathbb P}(\omega_{n+1}=T|\omega_1\omega_2\cdots\omega_n) := \tilde q = \frac{u-1-r}{u-d}.
\]
So $\omega_n$'s are independent of each other under $\widetilde {\mathbb P}$ and we have
\[
\widetilde {\mathbb P}(S_3=32)=\tilde p^3=\frac{1}{8}, \;\;
\widetilde {\mathbb P}(S_3=8)=3\tilde p^2\tilde q=\frac{3}{8}, \;\;
\widetilde {\mathbb P}(S_3=2)=3\tilde p\tilde q^2=\frac{3}{8}, \;\;
\widetilde {\mathbb P}(S_3=0.5)=\tilde q^3=\frac{1}{8}.
\]
\end{solution}

\noindent (ii) Compute $\widetilde {\mathbb E}S_1$, $\widetilde {\mathbb E}S_2$, and $\widetilde {\mathbb E}S_3$. What is the average rate of growth of the stock price under $\widetilde {\mathbb P}$?

\begin{solution}
\[
\begin{cases}
\widetilde {\mathbb E}[S_1]=8\widetilde {\mathbb P}(S_1=8)+2\widetilde
{\mathbb P}(S_1=2)=8\tilde p+2\tilde q=5 \\
\widetilde {\mathbb E}[S_2]=16\tilde p^2+4\cdot 2\tilde p\tilde q+1\cdot
\tilde q^2=6.25\\
\widetilde {\mathbb E}[S_3]=32\cdot \frac{1}{8}+8\cdot \frac{3}{8}+2\cdot \frac{3}{8}+0.5\cdot
\frac{1}{8}=7.8125.
\end{cases}
\]

So the average rates of growth of the stock
price under $\widetilde {\mathbb P}$ are, respectively:
\[
\begin{cases}
\tilde r_0=\frac{\widetilde {\mathbb E}[S_1]}{S_0} - 1=\frac{5}{4}-1=0.25, \\
\tilde r_1=\frac{\widetilde {\mathbb E}[S_2]}{\widetilde {\mathbb E}[S_1]}-1=\frac{6.25}{5}-1=0.25, \\
\tilde r_2=\frac{\widetilde {\mathbb E}[S_3]}{\widetilde {\mathbb E}[S_2]}-1=\frac{7.8125}{6.25}-1=0.25.
\end{cases}
\]
\end{solution}

\begin{remark}
An alternative solution is to use martingale property:
\[
\widetilde {\mathbb E}\left[\frac{S_n}{(1+r)^n}\right]=S_0 \Rightarrow \widetilde {\mathbb E}\left[S_n\right]=(1+r)^nS_0, \; \frac{\widetilde {\mathbb E}\left[S_n\right]}{\widetilde {\mathbb E}\left[S_{n-1}\right]}=(1+r).
\]
\end{remark}

\noindent (iii) Answer (i) and (ii) again under the actual probabilities $p=\frac{2}{3}$, $q=\frac{1}{3}$.

\begin{solution}${\mathbb P}(S_3=32)=(\frac{2}{3})^3=\frac{8}{27}$,
${\mathbb P}(S_3=8)=3\cdot (\frac{2}{3})^2 \cdot \frac{1}{3}=\frac{4}{9}$,
${\mathbb P}(S_3=2)=2\cdot \frac{1}{9}=\frac{2}{9}$, and
${\mathbb P}(S_3=0.5)=\frac{1}{27}$.

Accordingly, ${\mathbb E}[S_1]=6$, ${\mathbb E}[S_2]=9$ and ${\mathbb E}[S_3]=13.5$. So the
average rates of growth of the stock price under $\mathbb P$ are,
respectively: $r_0=\frac{6}{4}-1=0.5$, $r_1=\frac{9}{6}-1=0.5$, and
$r_2=\frac{13.5}{9}-1=0.5$.
\end{solution}

\begin{remark}
An alternative solution is to use Markov property: ${\mathbb E}_n[S_{n+1}]= S_n {\mathbb E}_n[S_{n+1}/S_n] = S_n (pu+qd)$.
\end{remark}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.3.} Show that a convex function of a martingale is a submartingale. In other words, let $M_0$, $M_1$, $\cdots$, $M_N$ be a martingale and let $\varphi$ be a convex function. Show that $\varphi(M_0)$, $\varphi(M_1)$, $\cdots$, $\varphi(M_N)$ is a submartingale.

 \begin{proof}Apply conditional Jensen's inequality, Theorem 2.3.2 (v). \end{proof}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.4.} Toss a coin repeatedly. Assume the probability of head on each toss is $\frac{1}{2}$, as is the probability of tail. Let $X_j=1$ if the $j$th toss results in a head and $X_j=-1$ if the $j$th toss results in a tail. Consider the stochastic process $M_0$, $M_1$, $M_2$, $\cdots$ defined by $M_0=0$ and
\[
M_n = \sum_{j=1}^n X_j, n \ge 1.
\]
This is called a {\it symmetric random walk}; with each head, it steps up one, and with each tail, it steps down one.

\smallskip

\noindent (i) Using the properties of Theorem 2.3.2, show that $M_0$, $M_1$, $M_2$, $\cdots$ is a martingale.

\begin{proof}
${\mathbb E}_n[M_{n+1}]=M_n+{\mathbb E}_n[X_{n+1}]=M_n+{\mathbb E}[X_{n+1}]=M_n$.
\end{proof}

\noindent (ii) Let $\sigma$ be a positive constant and, for $n\ge 0$, define
\[
S_n = e^{\sigma M_n} \left(\frac{2}{e^{\sigma}+e^{-\sigma}}\right)^n.
\]
Show that $S_0$, $S_1$, $S_2$, $\cdots$ is a martingale. Note that even though the symmetric random walk $M_n$ has no tendency to grow, the ``geometric symmetric random walk" $e^{\sigma M_n}$ does have a tendency to grow. This is the result of putting a martingale into the (convex) exponential function (see Exercise 2.3). In order to again have a martingale, we must ``discount" the geometric symmetric random walk, using the term $\frac{2}{e^{\sigma}+e^{-\sigma}}$ as the discount rate. This term is strictly less than one unless $\sigma=0$.

\begin{proof}
\[
{\mathbb E}_n \left[\frac{S_{n+1}}{S_n}\right]
={\mathbb E}_n \left[e^{\sigma X_{n+1}}\frac{2}{e^{\sigma}+e^{-\sigma}}\right]
=\frac{2}{e^{\sigma}+e^{-\sigma}} {\mathbb E}\left[e^{\sigma X_{n+1}}\right]
=1.
\]
\end{proof}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.5.} Let $M_0$, $M_1$, $M_2$, $\cdots$ be the symmetric random walk of Exercise 2.4, and define $I_0=0$ and
\[
I_n = \sum_{j=0}^{n-1} M_j (M_{j+1}-M_j), \; n=1,2,\cdots.
\]

\smallskip

\noindent (i) Show that
\[
I_n = \frac{1}{2}M_n^2 - \frac{n}{2}.
\]

\begin{proof}
\begin{eqnarray*}
2I_n
&=& 2\sum_{j=0}^{n-1}M_j(M_{j+1}-M_j) = 2\sum_{j=0}^{n-1}M_jM_{j+1}+M_n^2-\sum_{j=0}^{n-1}M_{j+1}^2-\sum_{j=0}^{n-1}M_j^2 \\
&=& M_n^2-\sum_{j=0}^{n-1}(M_{j+1}-M_j)^2 = M_n^2-\sum_{j=0}^{n-1}X_{j+1}^2 = M_n^2-n.
\end{eqnarray*}
\end{proof}
\begin{remark}
This is the discrete version of the integration-by-parts formula for stochastic integral:
\[
M_T^2 - M_0^2 = 2 \int_0^T M_t dM_t + [M, M]_T.
\]
with $I_T=\int_0^TM_tdM_t$.
\end{remark}

\noindent (ii) Let $n$ be an arbitrary nonnegative integer, and let $f(i)$ be an arbitrary function of a variable $i$. In terms of $n$ and $f$, define another function $g(i)$ satisfying
\[
{\mathbb E}_n[f(I_{n+1})] = g(I_n).
\]
Note that although the function $g(I_n)$ on the right-hand side of this equation may depend on $n$, the only random variable that may appear in its argument is $I_n$; the random variable $M_n$ may not appear. You will need to use the formula in part (i). The conclusion of part (ii) is that the process $I_0$, $I_1$, $I_2$, $\cdots$ is a Markov process.

\begin{solution}
\begin{eqnarray*}
& & {\mathbb E}_n[f(I_{n+1})] \\
&=& {\mathbb E}_n[f(I_n+M_n(M_{n+1}-M_n))] \\
&=& {\mathbb E}_n[f(I_n+M_nX_{n+1})] \\
&=& \frac{1}{2}[f(I_n+M_n)+f(I_n-M_n)] \\
&=& g(I_n),
\end{eqnarray*}
where $g(x)=\frac{1}{2}[f(x+\sqrt{2x+n})+f(x-\sqrt{2x+n})]$, since $\sqrt{2I_n+n}=|M_n|$.
\end{solution}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.6 (Discrete-time stochastic integral).} Suppose $M_0$, $M_1$, $\cdots$, $M_N$ is a martingale, and let $\Delta_0$, $\Delta_1$, $\cdots$, $\Delta_{N-1}$ be an adapted process. Define the {\it discrete-time stochastic integral} (sometimes called a {\it martingale transform}) $I_0$, $I_1$, $\cdots$, $I_N$ by setting $I_0=0$ and
\[
I_n = \sum_{j=0}^{n-1} \Delta_j (M_{j+1}-M_j), \; n=1, \cdots, N.
\]
Show that $I_0$, $I_1$, $\cdots$, $I_N$ is a martingale.

\begin{proof}${\mathbb E}_n[I_{n+1}-I_n]={\mathbb E}_n[\Delta_n(M_{n+1}-M_n)]=\Delta_n{\mathbb E}_n[M_{n+1}-M_n]=0$.\end{proof}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.7.} In a binomial model, give an example of a stochastic process that is a martingale but is not Markov.

\begin{solution}
We denote by $X_n$ the result of the $n$th coin toss:
\[
\begin{cases}
X_n = 1 & \mbox{if $\omega_n = H$} \\
X_n = -1 & \mbox{if $\omega_n = T$}
\end{cases}
\]
We also
suppose ${\mathbb P}(X=1)={\mathbb P}(X=-1)=\frac{1}{2}$. Define $S_1=X_1$ and
$S_{n+1}=S_{n}+b_n(X_1,\cdots,X_n)X_{n+1}$, where $b_n(\cdot)$ is a
bounded function on $\{-1,1\}^n$, to be determined later on. Clearly
$(S_n)_{n\ge 1}$ is an adapted stochastic process, and we can show
it is a martingale. Indeed,
\[
{\mathbb E}_n[S_{n+1}-S_n]=b_n(X_1,\cdots,X_n) {\mathbb E}_n[X_{n+1}]=0.
\]

For any arbitrary function $f$,
${\mathbb E}_n[f(S_{n+1})]=\frac{1}{2}[f(S_n+b_n(X_1,\cdots,X_n))+f(S_n-b_n(X_1,\cdots,X_n))]$.
Then intuitively, ${\mathbb E}_n[f(S_{n+1}]$ cannot be solely dependent upon
$S_n$ when $b_n$'s are properly chosen. Therefore in general,
$(S_n)_{n\ge 1}$ cannot be a Markov process.
\end{solution}

\begin{remark} If $X_n$ is regarded as the gain/loss of n-th bet in a
gambling game, then $S_n$ would be the wealth at time $n$. $b_n$ is
therefore the wager for the $(n+1)$th bet and is devised according to
past gambling results.
\end{remark}


\medskip

\noindent $\blacktriangleright$  {\bf Exercise 2.8.} Consider an $N$-period binomial model.

\smallskip

\noindent (i) Let $M_0$, $M_1$, $\cdots$, $M_N$ and $M_0'$, $M_1'$, $\cdots$, $M_N'$ be martingales under the risk-neutral measure $\widetilde {\mathbb P}$. Show that if $M_N=M_N'$ (for every possible outcome of the sequence of coin tosses), then, for each $n$ between $0$ and $N$, we have $M_n=M_n'$ (for every possible outcome of the sequence of coin tosses).

\begin{proof} Note $M_n= {\mathbb E}_n[M_N]={\mathbb E}_n[M_N']=M_n'$, $n=0, 1, \cdots, N$.\end{proof}

\noindent (ii) Let $V_N$ be the payoff at time $N$ of some derivative security. This is a random variable that can depend on all $N$ coin tosses. Define recursively $V_{N-1}$, $V_{N-2}$, $\cdots$, $V_0$ by the algorithm (1.2.16) of Chapter 1. Show that
\[
V_0, \frac{V_1}{1+r}, \cdots, \frac{V_{N-1}}{(1+r)^{N-1}}, \frac{V_N}{(1+r)^N}
\]
is a martingale under $\widetilde {\mathbb P}$.

\begin{proof} In the proof of Theorem 1.2.2, we proved by
induction that $X_n=V_n$ where $X_n$ is defined by (1.2.14) of
Chapter 1: $X_{n+1}=\Delta_nS_{n+1}+(1+r)(X_n-\Delta_nS_n)$. Since $\left\{\frac{X_n}{(1+r)^n}\right\}_{0\le
n\le N}$ is a martingale under $\widetilde {\mathbb P}$ (Theorem 2.4.5),
$\left\{\frac{V_n}{(1+r)^n}\right\}_{0\le n\le N}$ is also a martingale under
$\widetilde {\mathbb P}$.
\end{proof}

\noindent (iii) Using the risk-neutral pricing formula (2.4.11) of this chapter, define
\[
V_n' = \widetilde {\mathbb E}_n \left[\frac{V_N}{(1+r)^{N-n}}\right], n=0, 1, \cdots, N-1.
\]
Show that
\[
V_0', \frac{V_1'}{1+r}, \cdots, \frac{V_{N-1}'}{(1+r)^{N-1}}, \frac{V_N}{(1+r)^N}
\]
is a martingale.

\begin{proof}
This is obvious by iterated conditioning.
\end{proof}

\noindent (iv) Conclude that $V_n=V_n'$ for every $n$ (i.e., the algorithm (1.2.16) of Theorem 1.2.2 of Chapter 1 gives the same derivative security prices as the risk-neutral pricing formula (2.4.11) of Chapter 2).
\begin{proof} Combine (ii) and (iii), then use (i).
\end{proof}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.9 (Stochastic volatility, random interest rate).} Consider a two-period stochastic volatility, random interest rate model of the type described in Exercise 1.9 of Chapter 1. The stock prices and interest rates are shown in Figure 2.8.1.

\begin{figure}
\centering
% Set the overall layout of the tree
\tikzstyle{level 1}=[level distance=3.5cm, sibling distance=3.5cm]
\tikzstyle{level 2}=[level distance=3.5cm, sibling distance=2cm]

% Define styles for bags and leafs
\tikzstyle{bag} = [text width=5em, text centered]
\tikzstyle{end} = []

\begin{tikzpicture}[grow=right, sloped]
\node[bag] {$S_0=4$ $r_0=\frac{1}{4}$}
    child {
        node[bag] {$S_1(T)=2$ $r_1(T)=\frac{1}{2}$}
            child {
                node[end, label=right:
                    {$S_2(TT)=2$}] {}
                edge from parent
                node[above] {}
                node[below]  {}
            }
            child {
                node[end, label=right:
                    {$S_2(TH)=8$}] {}
                edge from parent
                node[above] {}
                node[below]  {}
            }
            edge from parent
            node[above] {}
            node[below]  {}
    }
    child {
        node[bag] {$S_1(H)=8$ $r_1(H)=\frac{1}{4}$}
        child {
                node[end, label=right:
                    {$S_2(HT)=8$}] {}
                edge from parent
                node[above] {}
                node[below]  {}
            }
            child {
                node[end, label=right:
                    {$S_2(HH)=12$}] {}
                edge from parent
                node[above] {}
                node[below]  {}
            }
        edge from parent
            node[above] {}
            node[below]  {}
    };
\end{tikzpicture}
\caption{{\bf Fig. 2.8.1.} A stochastic volatility, random interest rate model.}
\end{figure}

\smallskip

\noindent (i) Determine risk-neutral probabilities
\[
\widetilde {\mathbb P}(HH), \widetilde {\mathbb P}(HT), \widetilde {\mathbb P}(TH), \widetilde {\mathbb P}(TT),
\]
such that the time-zero value of an option that pays off $V_2$ at time two is given by the risk-neutral pricing formula
\[
V_0 = \widetilde {\mathbb E} \left[\frac{V_2}{(1+r_0)(1+r_1)}\right].
\]

\begin{solution}
\begin{eqnarray*}
& & u_0=\frac{S_1(H)}{S_0}=2, \; d_0=\frac{S_1(H)}{S_0}=\frac{1}{2}, \\
& & u_1(H)=\frac{S_2(HH)}{S_1(H)}=1.5, \; d_1(H)=\frac{S_2(HT)}{S_1(H)}=1, \\
& & u_1(T)=\frac{S_2(TH)}{S_1(T)}=4, \; d_1(T)=\frac{S_2(TT)}{S_1(T)}=1.
\end{eqnarray*}
Therefore
\begin{eqnarray*}
& & \tilde p_0=\frac{1+r_0-d_0}{u_0-d_0}=\frac{1}{2},\; \tilde q_0=1-\tilde p_0=\frac{1}{2} \\
& & \tilde p_1(H)=\frac{1+r_1(H)-d_1(H)}{u_1(H)-d_1(H)}=\frac{1}{2}, \; \tilde q_1(H)=1-\tilde p_1(H) = \frac{1}{2}, \\
& & \tilde p_1(T)=\frac{1+r_1(T)-d_1(T)}{u_1(T)-d_1(T)}=\frac{1}{6}, \; \tilde q_1(T)=1-\tilde p_1(T)=\frac{5}{6}.
\end{eqnarray*}
and
\begin{eqnarray*}
\widetilde {\mathbb P}(HH)=\tilde p_0\tilde
p_1(H)=\frac{1}{4}, \;
\widetilde {\mathbb P}(HT)=\tilde p_0\tilde
q_1(H)=\frac{1}{4}, \;
\widetilde {\mathbb P}(TH)=\tilde q_0\tilde
p_1(T)=\frac{1}{12}, \;
\widetilde {\mathbb P}(TT)=\tilde q_0\tilde
q_1(T)=\frac{5}{12}.
\end{eqnarray*}

The proofs of Theorem 2.4.4, Theorem 2.4.5 and Theorem 2.4.7 still
work for the random interest rate model, with proper modifications
(i.e. $\widetilde {\mathbb P}$ would be constructed according to the family of {\it
conditional probabilities} $\widetilde
{\mathbb P}(\omega_{n+1}=H|\omega_1,\cdots,\omega_n):=\tilde p_n$ and
$\widetilde {\mathbb P}(\omega_{n+1}=T|\omega_1,\cdots,\omega_n):=\tilde
q_n$. See Footnote \ref{footnote:Ionescu-Tulcea} for comments.). So the time-zero value of an option
that pays off $V_2$ at time two is given by the risk-neutral pricing
formula $V_0=\widetilde {\mathbb E}\left[\frac{V_2}{(1+r_0)(1+r_1)}\right]$.
\end{solution}

\noindent (ii) Let $V_2=(S_2-7)^+$. Compute $V_0$, $V_1(H)$, and $V_1(T)$.

\begin{solution} $V_2(HH)=5$, $V_2(HT)=1$, $V_2(TH)=1$ and
$V_2(TT)=0$. So
\begin{eqnarray*}
V_1(H) &=& \frac{\tilde p_1(H)V_2(HH)+\tilde
q_1(H)V_2(HT)}{1+r_1(H)}=2.4 \\
V_1(T) &=& \frac{\tilde
p_1(T)V_2(TH)+\tilde q_1(T)V_2(TT)}{1+r_1(T)}=\frac{1}{9} \\
V_0 &=& \frac{\tilde p_0V_1(H)+\tilde q_0V_1(T)}{1+r_0} =
1.00444.
\end{eqnarray*}
\end{solution}

\noindent (iii) Suppose an agent sells the option in (ii) for $V_0$ at time zero. compute the position $\Delta_0$ she should take in the stock at time zero so that at time one, regardless of whether the first coin toss results in head or tail, the value of her portfolio is $V_1$.

\begin{solution}$\Delta_0=\frac{V_1(H)-V_1(T)}{S_1(H)-S_1(T)}=\frac{2.4-\frac{1}{9}}{8-2}=0.4-\frac{1}{54}\approx
0.3815$.\end{solution}

\noindent (iv) Suppose in (iii) that the first coin toss results in head. What position $\Delta_1(H)$ should the agent now take in the stock be sure that, regardless of whether the second coin toss results in head or tail, the value of her portfolio at time two will be $(S_2-7)^+$?

\begin{solution}
$\Delta_1(H)=\frac{V_2(HH)-V_2(HT)}{S_2(HH)-S_2(HT)}=\frac{5-1}{12-8}=1$.
\end{solution}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.10 (Dividend-paying stock).}\footnote{Compare this problem with \S5.5 of Shreve \cite{Shreve04b}.} We consider a binomial asset pricing model as in Chapter 1, except that, after each movement in the stock price, a dividend is paid and the stock price is reduced accordingly. To describe this in equations, we define
\[
Y_{n+1}(\omega_1\cdots\omega_n\omega_{n+1}) =
\begin{cases}
u, & \mbox{if $\omega_{n+1}=H$,}\\
d, & \mbox{if $\omega_{n+1}=T$.}
\end{cases}
\]
Note that $Y_{n+1}$ depends only on the $(n+1)$st coin toss. In the binomial model of Chapter 1, $Y_{n+1}S_n$ was the stock price at time $n+1$. In the dividend-paying model considered here, we have a random variable $A_{n+1}(\omega_1\cdots\omega_n\omega_{n+1})$, taking values in $(0,1)$, and the dividend paid at time $n+1$ is $A_{n+1}Y_{n+1}S_n$. After the dividend is paid, the stock price at time $n+1$ is
\[
S_{n+1}=(1-A_{n+1})Y_{n+1}S_n.
\]

An agent who begins with initial capital $X_0$ and at each time $n$ takes a position of $\Delta_n$ shares of stock, where $\Delta_n$ depends only on the first $n$ coin tosses, has a portfolio value governed by the wealth equation (see (2.4.6))\footnote{Note the assumption of the equation is that the dividends are reinvested. Also note the equation can be written as
\[
X_{n+1}-X_n = \Delta_n(S_{n+1}-S_n) + r(X_n-\Delta_nS_n) + \Delta_nA_{n+1}Y_{n+1}S_n,
\]
which gives three sources of wealth change: change of stock price, interest earned from money market account, and stock dividends.}
\begin{equation}
X_{n+1}=\Delta_n S_{n+1} + (1+r)(X_n-\Delta_nS_n) + \Delta_nA_{n+1}Y_{n+1}S_n
= \Delta_n Y_{n+1}S_n + (1+r)(X_n-\Delta_nS_n). \tag{2.8.2}
\end{equation}

\smallskip

\noindent (i) Show that the discounted wealth process is a martingale under the risk-neutral measure (i.e., Theorem 2.4.5 still holds for the wealth process (2.8.2)). As usual, the risk-neutral measure is still defined by the equations
\[
\tilde p = \frac{1+r-d}{u-d}, \; \tilde q = \frac{u-1-r}{u-d}.
\]

\begin{proof}
\begin{eqnarray*}
\widetilde
{\mathbb E}_n \left[\frac{X_{n+1}}{(1+r)^{n+1}}\right]
&=&\widetilde
{\mathbb E}_n \left[\frac{\Delta_nY_{n+1}S_n}{(1+r)^{n+1}}+\frac{(1+r)(X_n-\Delta_nS_n)}{(1+r)^{n+1}}\right]\\
&=&\frac{\Delta_nS_n}{(1+r)^{n+1}}\widetilde
{\mathbb E}_n[Y_{n+1}]+\frac{X_n-\Delta_nS_n}{(1+r)^{n}}\\
&=&\frac{\Delta_nS_n}{(1+r)^{n+1}}(u\tilde
p+d\tilde
q)+\frac{X_n-\Delta_nS_n}{(1+r)^{n}}\\
&=&\frac{\Delta_nS_n+X_n-\Delta_nS_n}{(1+r)^n}=\frac{X_n}{(1+r)^n}.
\end{eqnarray*}
\end{proof}

\noindent (ii) Show that the risk-neutral pricing formula still applies (i.e., Theorem 2.4.7 holds for the dividend-paying model).

\begin{proof} From (2.8.2), we have
\begin{eqnarray*}
\left\{
\begin{matrix}
\Delta_nuS_n+(1+r)(X_n-\Delta_nS_n)=X_{n+1}(H)\\
\Delta_ndS_n+(1+r)(X_n-\Delta_nS_n)=X_{n+1}(T).
\end{matrix}
\right.
\end{eqnarray*}
So
\[
\Delta_n=\frac{X_{n+1}(H)-X_{n+1}(T)}{uS_n-dS_n},\;
X_n=\widetilde {\mathbb E}_n\left[\frac{X_{n+1}}{1+r}\right].
\]
To make the portfolio
replicate the payoff at time $N$, we must have $X_N=V_N$. So
$X_n=\widetilde {\mathbb E}_n \left[\frac{X_N}{(1+r)^{N-n}}\right]=\widetilde
{\mathbb E}_n \left[\frac{V_N}{(1+r)^{N-n}} \right]$. Since $(X_n)_{0\le n\le N}$ is the
value process of the unique replicating portfolio (uniqueness is
guaranteed by the uniqueness of the solution to the above linear
equations), the no-arbitrage price of $V_N$ at time $n$ is
$V_n=X_n=\widetilde {\mathbb E}_n \left[\frac{V_N}{(1+r)^{N-n}}\right]$.
\end{proof}

\noindent (iii) Show that the discounted stock price is not a martingale under the risk-neutral measure (i.e., Theorem 2.4.4 no longer holds). However, if $A_{n+1}$ is a constant $a\in (0,1)$, regardless of the value of $n$ and the outcome of the coin tossing $\omega_1\cdots\omega_{n+1}$, then $\frac{S_n}{(1-a)^n(1+r)^n}$ is a martingale under the risk-neutral measure.

\begin{proof}\begin{eqnarray*}
& & \widetilde {\mathbb E}_n \left[\frac{S_{n+1}}{(1+r)^{n+1}}\right] \\
&=&\frac{1}{(1+r)^{n+1}}\widetilde {\mathbb E}_n[(1-A_{n+1})Y_{n+1}S_n]\\
&=&\frac{S_n}{(1+r)^{n+1}}\left\{\tilde p[1-A_{n+1}(\omega_1\cdots\omega_nH)]u+\tilde
q[1-A_{n+1}(\omega_1\cdots\omega_nT)]d\right\}\\
&<&\frac{S_n}{(1+r)^{n+1}}[\tilde pu+\tilde qd]\\
&=&\frac{S_n}{(1+r)^n}.
\end{eqnarray*}
If $A_{n+1}$ is a constant $a$, then
\[
\widetilde
{\mathbb E}_n\left[\frac{S_{n+1}}{(1+r)^{n+1}}\right]=\frac{S_n}{(1+r)^{n+1}}(1-a)(\tilde
p u+\tilde q d)=\frac{S_n}{(1+r)^n}(1-a).
\]
So $\widetilde
{\mathbb E}_n \left[\frac{S_{n+1}}{(1+r)^{n+1}(1-a)^{n+1}}\right]=\frac{S_n}{(1+r)^n(1-a)^n}$, which implies $\frac{S_n}{(1-a)^n(1+r)^n}$ is a martingale under the risk-neutral measure.
\end{proof}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.11 (Put-call parity).} Consider a stock that pays no dividend in an $N$-period binomial model. A European call has payoff $C_N = (S_N - K)^+$ at time $N$. The price $C_n$ of this call at earlier times is given by the risk-neutral pricing formula (2.4.11):
\[
C_n = \widetilde {\mathbb E}_n \left[\frac{C_N}{(1+r)^{N-n}}\right], \; n = 0, 1, \cdots, N-1.
\]
Consider also a put with payoff $P_N = (K - S_N)^+$ at time $N$, whose price at earlier time is
\[
P_n = \widetilde {\mathbb E}_n \left[\frac{P_N}{(1+r)^{N-n}} \right], \; n=0, 1, \cdots, N-1.
\]
Finally, consider a {\it forward contract} to buy one share of stock at time $N$ for $K$ dollars. The price of this contract at time $N$ is $F_N = S_N-K$, and its price at earlier times is
\[
F_n = \widetilde {\mathbb E}_n \left[\frac{F_N}{(1+r)^{N-n}}\right], \; n = 0, 1, \cdots, N-1.
\]
(Note that, unlike the call, the forward contract requires that the stock be purchased at time $N$ for $K$ dollars and has a negative payoff if $S_N<K$.)

\smallskip

\noindent (i) If at time zero you buy a forward contract and a put, and hold them until expiration, explain why the payoff you receive is the same as the payoff of call; i.e., explain why $C_N = F_N +P_N$.

\begin{solution}
\[
F_N+P_N=S_N-K+(K-S_N)^+=
\begin{cases}
S_N-K+K-S_N & \mbox{if $K> S_N$} \\
S_N - K & \mbox{if $K\le S_N$}
\end{cases}
=(S_N-K)^+=C_N.
\]
\end{solution}

\noindent (ii) Using the risk-neutral pricing formulas given above for $C_n$, $P_n$, and $F_n$ and the linearity of conditional expectations, show that $C_n = F_n + P_n$ for every $n$.

\begin{proof} $C_n=\widetilde
{\mathbb E}_n[\frac{C_N}{(1+r)^{N-n}}]=\widetilde
{\mathbb E}_n[\frac{F_N}{(1+r)^{N-n}}]+\widetilde
{\mathbb E}_n[\frac{P_N}{(1+r)^{N-n}}]=F_n+P_n$.\end{proof}

\noindent (iii) Using the fact that the discounted stock price is a martingale under the risk-neutral measure, show that $F_0=S_0-\frac{K}{(1+r)^N}$.

\begin{proof} $F_0=\widetilde
{\mathbb E}[\frac{F_N}{(1+r)^N}]=\frac{1}{(1+r)^N}\widetilde
{\mathbb E}[S_N-K]=S_0-\frac{K}{(1+r)^N}$.
\end{proof}

\noindent (iv) Suppose you begin at time zero with $F_0$, buy one share of stock, borrowing money as necessary to do that, and make no further trades. Show that at time $N$ you have a portfolio valued at $F_N$. (This is called a {\it static replication} of the forward contract. If you sell the forward contract for $F_0$ at time zero, you can use this static replication to hedge your short position in the forward contract.)

\begin{proof} At time zero, the trader has $F_0-S_0$ in money
market account and one share of stock. At time $N$, the trader has a
wealth of $(F_0-S_0)(1+r)^N+S_N=-K+S_N=F_N$.
\end{proof}

\noindent (v) The {\it forward price} of the stock at time zero is defined to be that value of $K$ that causes the forward contract to have price zero at time zero. The forward price in this model is $(1+r)^NS_0$. Show that, at time zero, the price of a call struck at the forward price is the same as the price of a put struck at the forward price. This fact is called {\it put-call parity}.

\begin{proof} By (iii), $F_0=S_0-\frac{(1+r)^NS_0}{(1+r)^N}=0$. So by (ii), $C_0=F_0+P_0=P_0$.
\end{proof}
\begin{remark}
The {\it forward price} can be easily determined by the static replication: to hedge a short position in forward contract, we can buy and hold one share of stock until time $N$. The cost of this strategy is $S_0$ at time zero and $S_0(1+r)^N$ at time $N$. At time $N$, we receive $K$ dollars for compensation. So we must have $K=S_0(1+r)^N$ to eliminate arbitrage.
\end{remark}

\noindent (vi) If we choose $K=(1+r)^NS_0$, we just saw in (v) that $C_0=P_0$. Do we have $C_n=P_n$ for every $n$?

\begin{proof}By (ii), $C_n=P_n$ if and only if $F_n=0$. Note
$F_n=\widetilde
{\mathbb E}_n[\frac{S_N-K}{(1+r)^{N-n}}]=S_n-\frac{(1+r)^NS_0}{(1+r)^{N-n}}=S_n-S_0(1+r)^n$.
So $F_n$ is zero if and only if $n=0$ or $r=0$.
\end{proof}

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.12 (Chooser option).} Let $1\le m \le N-1$ and $K>0$ be given. A {\it chooser option} is a contract sold at time zero that confers on its owner the right to receive either a call or a put at time $m$. The owner of the chooser may wait until time $m$ before choosing. The call or put chosen expires at time $N$ with strike $K$. Show that the time-zero price of a chooser option is the sum of the time-zero price of a put, expiring at time $N$ and having strike price $K$, and a call, expiring at time $m$ and having strike price $\frac{K}{(1+r)^{N-m}}$. (Hint: Use put-call parity (Exercise 2.11).)

\begin{proof} First, the no-arbitrage price of the
chooser option at time $m$ must be $\max(C_m,P_m)$, where
\[
C_m=\widetilde {\mathbb E}_m\left[\frac{(S_N-K)^+}{(1+r)^{N-m}}\right]
\;\mbox{and}\;
P_m=\widetilde {\mathbb E}_m\left[\frac{(K-S_N)^+}{(1+r)^{N-m}}\right].
\]
That is, $C_m$ is the no-arbitrage price of a call option at time $m$
and $P_m$ is the no-arbitrage price of a put option at time $m$. Both
of the call and the put have maturity date $N$ and strike price $K$. Suppose the
market is liquid, then the chooser option is equivalent to receiving
a payoff of $\max(C_m,P_m)$ at time $m$. Therefore, its no-arbitrage price at time $0$ should be $\widetilde
{\mathbb E}[\frac{\max(C_m,P_m)}{(1+r)^m}]$.

By the put-call parity, $C_m=S_m-\frac{K}{(1+r)^{N-m}}+P_m$. So $\max(C_m,P_m)=P_m+ \left[S_m-\frac{K}{(1+r)^{N-m}}\right]^+$. Therefore,
the time-zero price of a chooser option is
\[
\widetilde {\mathbb E}\left[
\frac{P_m}{(1+r)^m}\right]+\widetilde
{\mathbb E}\left[\frac{\left(S_m-\frac{K}{(1+r)^{N-m}}\right)^+}{(1+r)^m}\right]=\widetilde
{\mathbb E}\left[\frac{(K-S_N)^+}{(1+r)^N}\right]+\widetilde
{\mathbb E}\left[\frac{\left(S_m-\frac{K}{(1+r)^{N-m}}\right)^+}{(1+r)^m}\right].
\]
The first term stands for the time-zero price of a put, expiring at
time $N$ and having strike price $K$, and the second term stands for
the time-zero price of a call, expiring at time $m$ and having
strike price $\frac{K}{(1+r)^{N-m}}$.
\end{proof}

%\begin{remark}
%It would be interesting to explicitly construct the portfolio process that replicates the payoff of a chooser option at time $N$.  Recall Theorem 1.2.2 provides the construction of a
%portfolio process $\Delta_0$, $\cdots$, $\Delta_{m-1}$, whose wealth $X_m$ at time
%$m$ is $\max(C_m,P_m)$. Starting from time $m$, for paths on which $C_m>P_m$, we can
%construct a portfolio process $\Delta_{m}'$, $\cdots$, $\Delta_{N-1}'$ whose
%payoff at time $N$ is $(S_N-K)^+$, while for paths on which $C_m \le P_m$, we can construct a portfolio process $\Delta_{m}''$, $\cdots$, $\Delta_{N-1}''$ whose payoff at time $N$ is $(K-S_N)^+$. By defining $(m\le k\le
%N-1)$
%\begin{eqnarray*}
%\Delta_k(\omega)=\left\{
%\begin{matrix}
%\Delta_k'(\omega)& \mbox{if $C_m(\omega)>P_m(\omega)$}\\
%\Delta_k''(\omega)& \mbox{if $C_m(\omega)<P_m(\omega)$},
%\end{matrix}
%\right.
%\end{eqnarray*}
%we get a portfolio process $(\Delta_n)_{0\le n\le N-1}$ whose payoff is the
%same as that of the chooser option. So the no-arbitrage price
%process of the chooser option must be equal to the value process of
%the replicating portfolio. In particular,
%\[
%V_0=X_0=\widetilde
%{\mathbb E}\left[\frac{X_m}{(1+r)^m}\right]=\widetilde {\mathbb E}\left[\frac{\max(C_m,P_m)}{(1+r)^m}\right].
%\]
%\end{remark}

\medskip

\pagebreak

\noindent  $\blacktriangleright$ {\bf Exercise 2.13 (Asian option).} Consider an $N$-period binomial model. An {\it Asian option} has a payoff based on the average stock price, i.e.,
\[
V_N = f \left( \frac{1}{N+1} \sum_{n=0}^N S_n \right),
\]
where the function $f$ is determined by the contractual details of the option.

(i) Define $Y_n = \sum_{k=0}^n S_k$ and use the Independence Lemma 2.5.3 to show that the two-dimensional process $\{S_n, Y_n \}$, $n=0,1,\cdots,N$ is Markov.

\begin{proof} Note the conditional distribution
\[
\widetilde {\mathbb P}(\omega_{n+1} = i | \omega_1,\cdots,\omega_n) =
\begin{cases}
\frac{1+r-d}{u-d} & \mbox{if $i=H$}\\
\frac{u-1-4}{u-d} & \mbox{if $i=T$}
\end{cases}
\]
is deterministic (i.e., independent of $\omega_1$, $\cdots$, $\omega_n$). So $\omega_n$'s are i.i.d. under $\widetilde {\mathbb P}$, as well as under ${\mathbb P}$. Hence, without loss of generality, we can work under
${\mathbb P}$. For any function $g$,
\begin{eqnarray*}
& & {\mathbb E}_n[g(S_{n+1},Y_{n+1})]\\
&=& {\mathbb E}_n \left[g \left(\frac{S_{n+1}}{S_n}S_n,Y_n+\frac{S_{n+1}}{S_n}S_n\right) \right] \\
&=&p g(uS_n,Y_n+uS_n)+q g(dS_n,Y_n+dS_n),
\end{eqnarray*}
which is a function of $(S_n,Y_n)$. This shows $(S_n,Y_n)_{0\le n\le N}$ is
Markov under ${\mathbb P}$.
\end{proof}

(ii) According to Theorem 2.5.8, the price $V_n$ of the Asian option at time $n$ is some function $v_n$ of $S_n$ and $Y_n$; i.e.
\[
V_n = v_n(S_n, Y_n), \; n=0, 1, \cdots, N.
\]
Give a formula for $v_N(s,y)$, and provide an algorithm for computing $v_n(s,y)$ in terms of $v_{n+1}$.

\begin{proof} Set $v_N(s,y)=f(\frac{y}{N+1})$. Then
$v_N(S_N,Y_N)=f\left(\frac{\sum_{n=0}^NS_n}{N+1}\right)=V_N$. Suppose $v_{n+1}$
is given, then
\begin{eqnarray*}
V_n &=& \widetilde {\mathbb E}_n \left[\frac{V_{n+1}}{1+r}\right] \\
&=& \widetilde
{\mathbb E}_n \left[\frac{v_{n+1}(S_{n+1},Y_{n+1})}{1+r}\right] \\
&=& \frac{1}{1+r}[\tilde
pv_{n+1}(uS_n,Y_n+uS_n)+\tilde
qv_{n+1}(dS_n,Y_n+dS_n)].
\end{eqnarray*}
So the formula for computing $v_n(s,y)$ in terms of $v_{n+1}$ is
\[
v_n(s,y)=\frac{\tilde p v_{n+1}(us,y+us)+\tilde
q v_{n+1}(ds,y+ds)}{1+r}.
\]
\end{proof}

\pagebreak

\medskip

\noindent  $\blacktriangleright$ {\bf Exercise 2.14 (Asian option continued).}
Consider an $N$-period binomial model, and let $M$ be a fixed number between $0$ and $N-1$. Consider an Asian
option whose payoff at time $N$ is
\[
V_n = f \left(\frac{1}{N-M} \sum_{n=M+1}^N S_n \right),
\]
where again the function $f$ is determined by the contractual details of the option.

(i) Define
\[
Y_n =
\begin{cases}
0, & \mbox{if $0\le n \le M$,} \\
\sum_{k=M+1}^n S_k, & \mbox{if $M+1\le n \le N$.}
\end{cases}
\]
Show that the two-dimensional process $(S_n, Y_n)$, $n=0, 1, \cdots, N$ is Markov (under the risk-neutral measure $\widetilde {\mathbb P}$).

\begin{proof}For $n\le M$, $(S_n,Y_n)=(S_n,0)$.
Since coin tosses $\omega_n$'s are i.i.d. under $\widetilde {\mathbb P}$,
$(S_n,Y_n)_{0\le n\le M}$ is Markov under $\widetilde P$. More
precisely, for any function $h$, $\widetilde
{\mathbb E}_n[h(S_{n+1})]=\tilde p h(uS_n)+\tilde q h(dS_n)$, for
$n=0,1,\cdots,M-1$.

For any function $g$ of two variables, we have
\[
\widetilde
{\mathbb E}_M[g(S_{M+1},Y_{M+1})]=\widetilde
{\mathbb E}_M[g(S_{M+1},S_{M+1})]=\tilde pg(uS_M,uS_M)+\tilde q
g(dS_M,dS_M).
\]
And for $n\ge M+1$,
\[
\widetilde
{\mathbb E}_n[g(S_{n+1},Y_{n+1})]=\widetilde
{\mathbb E}_n \left[g\left(\frac{S_{n+1}}{S_n}S_n,Y_n+\frac{S_{n+1}}{S_n}S_n\right)\right]=\tilde
pg(uS_n,Y_n+uS_n)+\tilde qg(dS_n,Y_n+dS_n).
\]
So $(S_n,Y_n)_{0\le n\le N}$ is Markov under $\widetilde {\mathbb P}$.
\end{proof}

(ii) According to Theorem 2.5.8, the price $V_n$ of the Asian option at time $n$ is some function $v_n$ of $S_n$ and $Y_n$, i.e.
\[
V_n = v_n(S_n,Y_n), \; n= 0, 1, \cdots, N.
\]
Of course, when $n\le M$, $Y_n$ is not random and does not need to be included in this function. Thus, for such $n$ we should seek a function $v_n$ of $S_n$ alone and have
\[
V_n =
\begin{cases}
v_n(S_n), & \mbox{if $0\le n \le M$,} \\
v_n(S_n, Y_n), & \mbox{if $M+1 \le n \le N$.}
\end{cases}
\]
Give a formula for $v_N(s,y)$, and provide an algorithm for computing $v_n$ in terms of $v_{n+1}$. Note that the algorithm is different for $n<M$ and $n>M$, and there is a separate transition formula for $v_M(s)$ in terms of $v_{M+1}(\cdot, \cdot)$.

\begin{proof} Set $v_N(s,y)=f(\frac{y}{N-M})$. Then
$v_N(S_N,Y_N)=f \left(\frac{\sum_{K=M+1}^NS_k}{N-M}\right)=V_N$.

Suppose $v_{n+1}$ is already given.

a) If $n>M$, then
\[
\widetilde
{\mathbb E}_n[v_{n+1}(S_{n+1},Y_{n+1})]=\tilde
pv_{n+1}(uS_n,Y_n+uS_n)+\tilde qv_{n+1}(dS_n,Y_n+dS_n).
\]
So
$v_n(s,y)=\tilde pv_{n+1}(us,y+us)+\tilde
qv_{n+1}(ds,y+ds)$.

b) If $n=M$, then
\[
\widetilde
{\mathbb E}_M[v_{M+1}(S_{M+1},Y_{M+1})]=\tilde
pv_{M+1}(uS_M,uS_M)+\tilde v_{n+1}(dS_M,dS_M).
\] So
$v_M(s)=\tilde p v_{M+1}(us,us)+\tilde qv_{M+1}(ds,ds)$.

c) If $n<M$, then
\[
\widetilde {\mathbb E}_n[v_{n+1}(S_{n+1})]=\tilde
pv_{n+1}(uS_n)+\tilde qv_{n+1}(dS_n).
\]
So $v_n(s)=\tilde
pv_{n+1}(us)+\tilde qv_{n+1}(ds)$.
\end{proof}

\section{State Prices}

$\bigstar$ {\bf Comments}:

\medskip

According to the {\it Fundamental Theorem of Asset Pricing} (FTAP), the no-arbitrage property is associated with the existence of a probability measure called {\it equivalent martingale measure} (EMM) and a positive process called  {\it num\'{e}raire}, such that the price processes of tradable assets discounted by the num\'{e}raire are martingales under the given probability measure (hence the name {\it martingale measure}).

Theorem 2.4.4 shows that the risk-neutral probability ${\mathbb P}$ defined by
\[
\tilde p = \frac{1+r-d}{u-d}, \; \tilde q = \frac{u-1-r}{u-d}
\]
and the value process of the money market account $\{(1+r)^n\}_{n=0}^N$ is such an EMM-num\'{e}raire pair. Theorem 3.2.7 shows that the actual probability ${\mathbb P}$  and the  inverse of the state price density process $\left\{\frac{1}{\zeta_n}\right\}_{n=0}^N$ is such an EMM-num\'{e}raire pair. Note num\'{e}raire does not have to the price process of a tradable asset, as the inverse of the state price density process is just an abstract stochastic process.

For a survey of the various formulations of the two Fundamental Theorem of Asset Pricing, we refer to Zeng \cite{Zeng09}.


\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 3.1.} Under the conditions of Theorem 3.1.1, show the following analogues of properties (i)-(iii) of that theorem:

\noindent (i') $\widetilde {\mathbb P}\left(\frac{1}{Z}>0\right) = 1$;

\noindent (ii') $\widetilde {\mathbb E}\frac{1}{Z}=1$;

\noindent (iii') for any random variable $Y$,
\[
{\mathbb E} Y = \widetilde {\mathbb E} \left[\frac{1}{Z} \cdot Y \right].
\]
In other words, $\frac{1}{Z}$ facilitates the switch from $\widetilde {\mathbb E}$ to ${\mathbb E}$ in the same way $Z$ facilitates the switch from ${\mathbb E}$ to $\widetilde {\mathbb E}$.

\begin{proof}
Define $\widetilde
Z(\omega):=\frac{{\mathbb P}(\omega)}{\widetilde
{\mathbb P}(\omega)}=\frac{1}{Z(\omega)}$. Apply Theorem 3.1.1 with ${\mathbb P}$,
$\widetilde {\mathbb P}$, $Z$ replaced by $\widetilde {\mathbb P}$, ${\mathbb P}$, $\widetilde Z$,
we get the analogues of properties (i)-(iii) of Theorem 3.1.1.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.2.} Let ${\mathbb P}$ be a probability measure on a finite probability space $\Omega$. In this problem, we allow the possibility that ${\mathbb P}(\omega)=0$ for some values of $\omega \in \Omega$. Let $Z$ be a random variable on $\Omega$ with the property that ${\mathbb P}(Z\ge 0)=1$ and ${\mathbb E}Z=1$. For $\omega \in \Omega$, define $\widetilde {\mathbb P}(\omega) = Z(\omega){\mathbb P}(\omega)$, and for events $A\subset \Omega$, define $\widetilde {\mathbb P}(A) = \sum_{\omega \in A} \widetilde {\mathbb P}(\omega)$. Show the following.

\smallskip

\noindent (i) $\widetilde {\mathbb P}$ is a probability measure; i.e., $\widetilde {\mathbb P}(\Omega)=1$.

\begin{proof}
$\widetilde
{\mathbb P}(\Omega)=\sum_{\omega\in \Omega}\widetilde
{\mathbb P}(\omega)=\sum_{\omega\in \Omega}Z(\omega){\mathbb P}(\omega)=E[Z]=1$.
\end{proof}

\noindent (ii) If $Y$ is a random variable, then $\widetilde {\mathbb E}Y = {\mathbb E}[ZY]$.

\begin{proof}
$\widetilde {\mathbb E}[Y]=\sum_{\omega\in
\Omega}Y(\omega)\widetilde {\mathbb P}(\omega)=\sum_{\omega\in
\Omega}Y(\omega)Z(\omega){\mathbb P}(\omega)={\mathbb E}[YZ]$.
\end{proof}

\noindent (iii) If $A$ is an event with ${\mathbb P}(A)=0$, then $\widetilde {\mathbb P}(A) = 0$.

\begin{proof}
$\widetilde {\mathbb P}(A)=\sum_{\omega\in
A}Z(\omega){\mathbb P}(\omega)$. Since ${\mathbb P}(A)=0$, ${\mathbb P}(\omega)=0$ for any
$\omega\in A$. So $\widetilde {\mathbb P}(A)=0$.
\end{proof}

\noindent (iv) Assume that ${\mathbb P}(Z > 0) = 1$. Show that if $A$ is an event with $\widetilde {\mathbb P}(A)=0$, then ${\mathbb P}(A)=0$.

\begin{proof}
If $\widetilde {\mathbb P}(A)=\sum_{\omega\in
A}Z(\omega) {\mathbb P}(\omega)=0$, by ${\mathbb P}(Z>0)=1$, we conclude ${\mathbb P}(\omega)=0$
for any $\omega\in A$. So ${\mathbb P}(A)=\sum_{\omega\in A}{\mathbb P}(\omega)=0$.
\end{proof}

When two probability measures agree which events have probability zero (i.e., ${\mathbb P}(A)=0$ if and only if $\widetilde {\mathbb P}(A)=0$), the measure are said to be {\it equivalent}. From (iii) and (iv) above, we see that ${\mathbb P}$ and $\widetilde {\mathbb P}$ are equivalent under the assumption that ${\mathbb P}(Z>0)=1$.

\noindent (v) Show that if ${\mathbb P}$ and $\widetilde {\mathbb P}$ are equivalent, then they agree which events have probability one (i.e., ${\mathbb P}(A)=1$ if and only if $\widetilde{\mathbb P}(A)=1$).

\begin{proof} ${\mathbb P}(A)=1\iff {\mathbb P}(A^c)=0\iff \widetilde {\mathbb P}(A^c)=0 \iff
\widetilde {\mathbb P}(A)=1$.
\end{proof}

\noindent (vi) Construct an example in which we have only ${\mathbb P}(Z\ge 0)=1$ and ${\mathbb P}$ and $\widetilde {\mathbb P}$ are not equivalent.

\begin{solution} Pick $\omega_0$ such that $1>{\mathbb P}(\omega_0)>0$. Define
\[
Z(\omega)=
\begin{cases}
0 & \mbox{if $\omega\ne\omega_0$}\\
\frac{1}{{\mathbb P}(\omega_0)} & \mbox{if $\omega=\omega_0$}.
\end{cases}
\]
Then ${\mathbb P}(Z\ge 0)=1$ and ${\mathbb E}[Z]=\frac{1}{{\mathbb P}(\omega_0)}\cdot {\mathbb P}(\omega_0)=1$.

Clearly
\[
\widetilde {\mathbb P}(\Omega\setminus\{\omega_0\})=\sum_{\omega\ne
\omega_0}Z(\omega){\mathbb P}(\omega)=0.
\]
But ${\mathbb P}(\Omega\setminus\{\omega_0\})=1-{\mathbb P}(\omega_0)>0$ since ${\mathbb P}(\omega_0)<1$.
Hence ${\mathbb P}$ and $\widetilde {\mathbb P}$ cannot be
equivalent.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.3.} Using the stock price model of Figure 3.1.1 and the actual probabilities $p=\frac{2}{3}$, $q=\frac{1}{3}$, define the estimates of $S_3$ at various times by
\[
M_n = {\mathbb E}_n[S_3], \; n = 0, 1, 2, 3.
\]
Fill in the values of $M_n$ in a tree like that of Figure 3.1.1. Verify that $M_n$, $n=0,1,2,3$, is a martingale.

\begin{solution} $M_3 = S_3$,
\begin{eqnarray*}
M_2(HH) &=& {\mathbb E}_2[S_3](HH) = p S_3(HHH) + q S_3(HHT) = \frac{2}{3} \cdot 32 + \frac{1}{3} \cdot 8 = 24 \\
M_2(HT) &=& {\mathbb E}_2[S_3](HT) = p S_3(HTH) + q S_3(HTT) = \frac{2}{3} \cdot 8 + \frac{1}{3} \cdot 2 = 6 \\
M_2(TH) &=& {\mathbb E}_2[S_3](TH) = p S_3(THH) + q S_3(THT) = \frac{2}{3}\cdot 8 + \frac{1}{3}\cdot 2 = 6 \\
M_2(TT) &=& {\mathbb E}_2[S_3](TT) = p S_3(TTH) + q S_3(TTT) = \frac{2}{3}\cdot 2 + \frac{1}{3}\cdot 0.5 = 1.5 \\
M_1(H) &=& {\mathbb E}_1[S_3](H) = p^2 S_3(HHH) + pq S_3(HHT) + qp S_3(HTH) + q^2 S_3(HTT) \\
       &=& \left(\frac{2}{3}\right)^2 \cdot 32 + \frac{2}{3}\cdot\frac{1}{3}\cdot(8+8)+\left(\frac{1}{3}\right)^2 \cdot 2 = 18 \\
M_1(T) &=& {\mathbb E}_1[S_3](T) = p^2 S_3(THH) + pq S_3(THT) + qp S_3(TTH) + q^2 S_3(TTT) \\
       &=& \left(\frac{2}{3}\right)^2 \cdot 8 + \frac{2}{3}\cdot\frac{1}{3}\cdot(2+2)+\left(\frac{1}{3}\right)^2 \cdot 0.5 = 4.5 \\
M_0 &=& {\mathbb E}_0[S_3] = \left(\frac{2}{3}\right)^3\cdot 32 + \left(\frac{2}{3}\right)^2\cdot \frac{1}{3} \cdot (8+8+8) + \frac{2}{3}\cdot\left(\frac{1}{3}\right)^2\cdot (2+2+2)+\left(\frac{1}{3}\right)^3\cdot 0.5 = 13.5\\
\end{eqnarray*}

\begin{figure}
\centering
 \begin{tikzpicture}[>=stealth,sloped]
    \matrix (tree) [%
      matrix of nodes,
      minimum size=.5cm,
      column sep=0.3cm,
      row sep=1cm,
    ]
    {
              &            &                     & $M_3(HHH)=32$                  \\
              &            & $M_2(HH)=24$        &                                \\
              & $M_1(H)=18$ &                     & $M_3(HHT)=M_3(HTH)=M_3(THH)=8$ \\
    $M_0=13.5$&            & $M_2(HT)=M_2(TH)=6$ &                                \\
              & $M_1(T)=4.5$ &                     & $M_3(HTT)=M_3(THT)=M_3(TTH)=2$ \\
              &            & $M_2(TT)=1.5$         &                                \\
              &            &                     & $M_3(TTT)=.50$                 \\
    };
    \draw[->] (tree-4-1) -- (tree-3-2);
    \draw[->] (tree-4-1) -- (tree-5-2);
    \draw[->] (tree-3-2) -- (tree-2-3);
    \draw[->] (tree-3-2) -- (tree-4-3);
    \draw[->] (tree-5-2) -- (tree-4-3);
    \draw[->] (tree-5-2) -- (tree-6-3);
    \draw[->] (tree-2-3) -- (tree-1-4);
    \draw[->] (tree-2-3) -- (tree-3-4);
    \draw[->] (tree-4-3) -- (tree-3-4);
    \draw[->] (tree-4-3) -- (tree-5-4);
    \draw[->] (tree-6-3) -- (tree-5-4);
    \draw[->] (tree-6-3) -- (tree-7-4);
  \end{tikzpicture}
\caption{{\bf Exercise 3.3.}}
\end{figure}

To verify $(M_n)_{n=0}^3$ is a martingale, we note $M_2 = {\mathbb E}_2[S_3] = {\mathbb E}_2[M_3]$, since $M_3=S_3$. Moreover, we have
\begin{eqnarray*}
{\mathbb E}_1[M_2](H) &=& pM_2(HH)+qM_2(HT)=\frac{2}{3}\cdot 24 + \frac{1}{3}\cdot 6 = 18 = M_1(H) \\
{\mathbb E}_1[M_2](T) &=& pM_2(TH)+qM_2(TT)=\frac{2}{3}\cdot 6 + \frac{1}{3}\cdot 1.5 = 4.5 = M_1(T) \\
{\mathbb E}_0[M_1](T) &=& pM_1(H)+qM_1(T)=\frac{2}{3}\cdot 18 + \frac{1}{3}\cdot 4.5 = 13.5 = M_0. \\
\end{eqnarray*}
Therefore, $(M_n)_{n=0}^3$ is a martingale.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.4.} This problem refers to the model of Example 3.1.2, whose Radon-Nikod\'{y}m process $Z_n$ appears in Figure 3.2.1.

\smallskip

\noindent (i) Compute the state price densities
\begin{eqnarray*}
& & \zeta_3(HHH), \\
& & \zeta_3(HHT)=\zeta_3(HTH)=\zeta_3(THH),\\
& & \zeta_3(HTT)=\zeta_3(THT)=\zeta_3(TTH),\\
& & \zeta_3(TTT)
\end{eqnarray*}
explicitly.

\begin{solution} By formula (3.1.5), we have
\begin{eqnarray*}
& & \zeta_3(HHH)=\frac{27}{64}\cdot \frac{1}{(1+0.25)^3}=0.216\\
& & \zeta_3(HHT)=\zeta_3(HTH)=\zeta_3(THH)=\frac{27}{32}\cdot \frac{1}{(1+0.25)^3}=0.432\\
& & \zeta_3(HTT)=\zeta_3(THT)=\zeta_3(TTH)=\frac{27}{16}\cdot \frac{1}{(1+0.25)^3}=0.864\\
& & \zeta_3(TTT)=\frac{27}{8}\cdot \frac{1}{(1+0.25)^3}=1.728.
\end{eqnarray*}
\end{solution}

\noindent (ii) Use the numbers computed in (i) in formula (3.1.10) to find the time-zero price of the Asian option of Exercise 1.8 of Chapter 1. You should get $v_0(4,4)$ computed in part (ii) of that exercise.

\begin{solution}
\begin{figure}[ht] % Place the float at the same point it occurs in the source text (however, not exactly at the spot)
\centering
% Set the overall layout of the tree
\tikzstyle{level 1}=[level distance=3.5cm, sibling distance=6cm, ->]
\tikzstyle{level 2}=[level distance=3.5cm, sibling distance=3cm, ->]
\tikzstyle{level 3}=[level distance=3.5cm, sibling distance=1.5cm, ->]

% Define styles for bags and leafs
\tikzstyle{bag} = [text width=7em, text centered]
\tikzstyle{end} = [text width=7em, text centered]

\begin{tikzpicture}[grow=right, sloped]
\node[bag]{$S_0=4$ $Y_0=4$}
    child {
        node[bag]{$S_1(T)=2$ $Y_1(T)=6$}
            child {
                node[bag]{$S_2(TT)=1$ $Y_2(TT)=7$ }
                    child {
                        node[end]{$S_3(TTT)=0.5$ $Y_3(TTT)=7.5$}
                    }
                    child {
                        node[end]{$S_3(TTH)=2$ $Y_3(TTH)=9$}
                    }
            }
            child {
                node[bag]{$S_2(TH)=4$ $Y_2(TH)=10$}
                    child {
                        node[end]{$S_3(THT)=2$ $Y_3(THT)=12$}
                    }
                    child {
                        node[end]{$S_3(THH)=8$ $Y_3(THH)=18$}
                    }
            }
    }
    child {
        node[bag]{$S_1(H)=8$ $Y_1(H)=12$ }
            child {
                node[bag]{$S_2(HT)=4$ $Y_2(HT)=16$ }
                    child {
                        node[end]{$S_3(HTT)=2$ $Y_3(HTT)=18$}
                    }
                    child {
                        node[end]{$S_3(HTH)=8$ $Y_3(HTH)=24$}
                    }
            }
            child {
                node[bag]{$S_2(HH)=16$ $Y_2(HH)=28$ }
                    child {
                        node[end]{$S_3(HHT)=8$ $Y_3(HHT)=36$}
                    }
                    child {
                        node[end]{$S_3(HHH)=32$ $Y_3(HHH)=60$}
                    }
            }
    }
;
\end{tikzpicture}
\caption{{\bf Exercise 1.8.} Asian option.}
\end{figure}

Recall in Example 3.1.2, we have $p=\frac{2}{3}$ and $q=\frac{1}{3}$. So, by formula (3.1.10), the time-zero price of the Asian option of Exercise 1.8 is
\begin{eqnarray*}
V_0 &=& {\mathbb E}\left[\zeta_3 \left(\frac{1}{4}Y_3-4\right)^+\right] = \sum_{\omega\in\Omega} \left(\frac{1}{4}Y_3(\omega)-4\right)^+\zeta(\omega){\mathbb P}(\omega)\\
&=&  \left(\frac{1}{4}\cdot 60-4\right)^+ \cdot 0.216 \cdot \left(\frac{2}{3}\right)^3
    +\left(\frac{1}{4}\cdot 36-4\right)^+ \cdot 0.432 \cdot \left(\frac{2}{3}\right)^2\cdot\left(\frac{1}{3}\right) + \\
& & \left(\frac{1}{4}\cdot 24-4\right)^+ \cdot 0.432 \cdot \left(\frac{2}{3}\right)^2\cdot\left(\frac{1}{3}\right)
    +\left(\frac{1}{4}\cdot 18-4\right)^+ \cdot 0.864 \cdot \left(\frac{1}{3}\right)^2\cdot\left(\frac{2}{3}\right)
    + \\
& & \left(\frac{1}{4}\cdot 18-4\right)^+ \cdot 0.432 \cdot \left(\frac{2}{3}\right)^2\cdot\left(\frac{1}{3}\right) \\
&=& 1.216.
\end{eqnarray*}
\end{solution}

\noindent (iii) Compute also the state price densities $\zeta_2(HT)=\zeta_2(TH)$.
\begin{solution}
We note $\zeta_2 = \frac{Z_2}{(1+r)^2} = {\mathbb E}_2\left[\frac{Z}{(1+r)^3}\right]\cdot (1+r)$.
By formula (3.1.5),
\begin{eqnarray*}
& & \zeta_2(HT) = \frac{1}{(1+r)^2}[pZ(HTH)+qZ(HTT)] = \frac{\frac{2}{3}\cdot\frac{27}{32} + \frac{1}{3}\cdot \frac{27}{16}}{1.25^2} = 0.72 \\
& & \zeta_2(TH)=\frac{1}{(1+r)^2}[pZ(THH)+qZ(THT)] = \frac{1}{(1+r)^2}[pZ(HTH)+qZ(HTT)] = 0.72.
\end{eqnarray*}
\end{solution}

\noindent (iv) Use the risk-neutral pricing formula (3.2.6) in the form
\begin{eqnarray*}
& & V_2(HT)=\frac{1}{\zeta_2(HT)} {\mathbb E}_2[\zeta_3V_3](HT), \\
& & V_2(TH)=\frac{1}{\zeta_2(TH)} {\mathbb E}_2[\zeta_3V_3](TH)
\end{eqnarray*}
to compute $V_2(HT)$ and $V_2(TH)$. You should get $V_2(HT)=v_2(4,16)$ and $V_2(TH)=v_2(4,10)$, where $v_2(s,y)$ was computed in part (ii) of Exercise 1.8 of Chapter 1. Note that $V_2(HT)\ne V_2(TH)$.

\begin{solution}
By Exercise 1.8 of Chapter 1, $V_2(HT)=v_2(4,16)=1$, $V_2(TH)=v_2(4,10)=0.2$. Meanwhile, we have
\begin{eqnarray*}
V_2(HT) &=& \frac{1}{\zeta_2(HT)} {\mathbb E}_2[\zeta_3V_3](HT) = \frac{1}{\zeta_2(HT)} [p\zeta_3(HTH)V_3(HTH)+q\zeta_3(HTT)V_3(HTT)] \\
&=& \frac{1}{0.72} \left[\frac{2}{3} \cdot 0.432 \cdot \left(\frac{1}{4}\cdot 24 -4\right)^+ + \frac{1}{3} \cdot 0.864 \cdot \left(\frac{1}{4}\cdot 18 -4\right)^+ \right] \\
&=&1,
\end{eqnarray*}
and
\begin{eqnarray*}
V_2(TH) &=& \frac{1}{\zeta_2(TH)} {\mathbb E}_2[\zeta_3V_3](TH) = \frac{1}{\zeta_2(TH)} [p\zeta_3(THH)V_3(THH)+q\zeta_3(THT)V_3(THT)] \\
&=& \frac{1}{0.72} \left[\frac{2}{3} \cdot 0.432 \cdot \left(\frac{1}{4}\cdot 18 -4\right)^+ + \frac{1}{3} \cdot 0.864 \cdot \left(\frac{1}{4}\cdot 12 -4\right)^+ \right] \\
&=&0.2.
\end{eqnarray*}
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 3.5 (Stochastic volatility, random interest rate).} Consider the model of Exercise 2.9 of Chapter 2. Assume that the actual probability measure is
\[
{\mathbb P}(HH) = \frac{4}{9}, \; {\mathbb P}(HT) = \frac{2}{9}, \; {\mathbb P}(TH)=\frac{2}{9}, \; {\mathbb P}(TT) = \frac{1}{9}.
\]
The risk-neutral measure was computed in Exercise 2.9 of Chapter 2.

\smallskip

\noindent (i) Compute the Radon-Nikod\'{y}m derivative $Z(HH)$, $Z(HT)$, $Z(TH)$, and $Z(TT)$ of $\widetilde {\mathbb P}$ with respect to ${\mathbb P}$.
\begin{solution} In Exercise 2.9 of Chapter 2, we have
\[
\widetilde {\mathbb P}(HH)=\frac{1}{4}, \;
\widetilde {\mathbb P}(HT)=\frac{1}{4}, \;
\widetilde {\mathbb P}(TH)=\frac{1}{12}, \;
\widetilde {\mathbb P}(TT)=\frac{5}{12}.
\]
Therefore
\begin{eqnarray*}
& & Z(HH)=\frac{\widetilde {\mathbb P}(HH)}{{\mathbb P}(HH)}=\frac{9}{16} \\
& & Z(HT)=\frac{\widetilde {\mathbb P}(HT)}{{\mathbb P}(HT)}=\frac{9}{8} \\
& & Z(TH)=\frac{\widetilde {\mathbb P}(TH)}{{\mathbb P}(TH)}=\frac{3}{8} \\
& & Z(TT)=\frac{\widetilde {\mathbb P}(TT)}{{\mathbb P}(TT)}=\frac{15}{4}.
\end{eqnarray*}
\end{solution}

\noindent (ii) The Radon-Nikod\'{y}m derivative process $Z_0$, $Z_1$, $Z_2$ satisfies $Z_2=Z$. Compute $Z_1(H)$, $Z_1(T)$, and $Z_0$. Note that $Z_0={\mathbb E}Z=1$.
\begin{solution}
From the specification of ${\mathbb P}(HH)$, ${\mathbb P}(HT)$, ${\mathbb P}(TH)$, and ${\mathbb P}(TT)$, we can conclude $\omega_1$ and $\omega_2$ are i.i.d. under ${\mathbb P}$ , with $p:={\mathbb P}(\omega_1=H)={\mathbb P}(\omega_2=H)=\frac{2}{3}$ and $q:={\mathbb P}(\omega_1=T)={\mathbb P}(\omega_2=T)=\frac{1}{3}$. Therefore,
\begin{eqnarray*}
& & Z_1(H) = {\mathbb E}_1[Z_2](H) = Z_2(HH)p+Z_2(HT)q=\frac{9}{16}\cdot \frac{2}{3}+\frac{9}{8}\cdot\frac{1}{3}=\frac{3}{4}, \\
& & Z_1(T) = {\mathbb E}_1[Z_2](T) = Z_2(TH)p+Z_2(TT)q=\frac{3}{8}\cdot \frac{2}{3}+\frac{15}{4}\cdot\frac{1}{3}=\frac{3}{2}, \\
& & Z_0 = {\mathbb E}[Z_1] = pZ_1(H)+qZ_1(T)= \frac{2}{3}\cdot \frac{3}{4}+\frac{1}{3}\cdot\frac{3}{2}=1.
\end{eqnarray*}
\end{solution}

\noindent (iii) The version of the risk-neutral pricing formula (3.2.6) appropriate for this model, which does not use the risk-neutral measure, is
\begin{eqnarray*}
V_1(H) &=& \frac{1+r_0}{Z_1(H)} {\mathbb E}_1 \left[\frac{Z_2}{(1+r_0)(1+r_1)}V_2\right](H) \\
&=& \frac{1}{Z_1(H)(1+r_1(H))}{\mathbb E}_1[Z_2V_2](H), \\
V_1(T) &=& \frac{1+r_0}{Z_1(H)} {\mathbb E}_1 \left[\frac{Z_2}{(1+r_0)(1+r_1)}V_2\right](T) \\
&=& \frac{1}{Z_1(H)(1+r_1(H))}{\mathbb E}_1[Z_2V_2](T),  \\
V_0 &=& {\mathbb E} \left[\frac{Z_2}{(1+r_0)(1+r_1)}V_2\right].
\end{eqnarray*}
Use this formula to compute $V_1(H)$, $V_1(T)$, and $V_0$ when $V_2=(S_2-7)^+$. Compare the result with your answers in Exercise 2.9(ii) of Chapter 2.\footnote{The textbook said ``Exercise 2.6" by mistake.}

\begin{solution}
\[
V_1(H)
=\frac{Z_2(HH)V_2(HH)p+Z_2(HT)V_2(HT)q}{Z_1(H)(1+r_1(H))}
=\frac{\frac{9}{16} \cdot (12-7)^+ \cdot \frac{2}{3} + \frac{15}{4} \cdot (8-7)^+ \cdot \frac{1}{3}}{\frac{3}{4} \cdot \left(1+\frac{1}{4}\right)}
=2.4,
\]
\[
V_1(T)=\frac{Z_2(TH)V_2(TH)p+Z_2(TT)V_2(TT)q}{Z_1(T)(1+r_1(T))}=\frac{\frac{3}{8}\cdot(8-7)^+\cdot\frac{2}{3}+\frac{15}{4}\cdot(2-7)^+\cdot\frac{1}{3}}{\frac{3}{2}\cdot \left(1+\frac{1}{2}\right)}=\frac{1}{9},
\]and
\begin{eqnarray*}
V_0
&=&\frac{Z_2(HH)V_2(HH)}{(1+r_0)(1+r_1(H))}{\mathbb P}(HH)+\frac{Z_2(HT)V_2(HT)}{(1+r_0)(1+r_1(H))}{\mathbb P}(HT)+\frac{Z_2(TH)V_2(TH)}{(1+r_0)(1+r_1(T))}{\mathbb P}(TH)+0 \\
&=&\frac{\frac{9}{16}\cdot (12-7)^+\cdot\frac{4}{9}}{(1+\frac{1}{4})(1+\frac{1}{4})}+\frac{\frac{9}{8}\cdot(8-7)^+\cdot\frac{2}{9}}{(1+\frac{1}{4})(1+\frac{1}{4})}+\frac{\frac{3}{8}\cdot(8-7)^+\cdot\frac{2}{9}}{(1+\frac{1}{4})(1+\frac{1}{2})} \\
&=&
1.00444.
\end{eqnarray*}
\end{solution}


\noindent $\blacktriangleright$ {\bf Exercise 3.6.} Consider Problem 3.3.1 in an $N$-period binomial model with the utility function $U(x)=\ln x$. Show that the optimal wealth process corresponding to the optimal portfolio process is given by $X_n=\frac{X_0}{\zeta_n}$, $n=0,1,\cdots,N$, where $\zeta_n$ is the state price density process defined in (3.2.7).

\begin{proof} $U'(x)=\frac{1}{x}$, so
$I(x)=\frac{1}{x}$. (3.3.26) gives
${\mathbb E} \left[\frac{Z}{(1+r)^N}\frac{(1+r)^N}{\lambda Z}\right]=X_0$. So
$\lambda=\frac{1}{X_0}$. By (3.3.25), we have
$X_N=\frac{(1+r)^N}{\lambda Z}=\frac{X_0}{Z}(1+r)^N$. Hence
\begin{eqnarray*}
X_n = \widetilde {\mathbb E}_n \left[\frac{X_N}{(1+r)^{N-n}} \right] = \widetilde
{\mathbb E}_n \left[\frac{X_0(1+r)^n}{Z} \right] = X_0(1+r)^n\widetilde
{\mathbb E}_n \left[\frac{1}{Z}\right] = X_0(1+r)^n\frac{1}{Z_n}{\mathbb E}_n\left[Z\cdot
\frac{1}{Z}\right]=\frac{X_0}{\zeta_n},
\end{eqnarray*}
where the second to last ``=" comes from Lemma 3.2.6.
\end{proof}

\noindent $\blacktriangleright$ {\bf Exercise 3.7.} Consider Problem 3.3.1 in an $N$-period binomial model with the utility function $U(x)=\frac{1}{p}x^p$, where $p<1$, $p\ne 0$. Show that the optimal wealth at time $N$ is
\[
X_N = \frac{X_0(1+r)^NZ^{\frac{1}{p-1}}}{{\mathbb E}\left[Z^{\frac{p}{p-1}}\right]},
\]
where $Z$ is the Radon-Nikod\'{y}m derivative of $\widetilde {\mathbb P}$ with respect to ${\mathbb P}$.

\begin{proof} $U'(x)=x^{p-1}$, so
$I(x)=x^{\frac{1}{p-1}}$. By (3.3.26), we have
${\mathbb E}\left[\frac{Z}{(1+r)^N}\left(\frac{\lambda
Z}{(1+r)^N}\right)^{\frac{1}{p-1}}\right]=X_0$. Solve it for $\lambda$, we get
\[
\lambda=\left(\frac{X_0}{{\mathbb E}\left[\frac{Z^{\frac{p}{p-1}}}{(1+r)^{
\frac{Np}{p-1}}}\right]}\right)^{p-1}=\frac{X_0^{p-1}(1+r)^{Np}}{\left({\mathbb E}\left[Z^{\frac{p}{p-1}}\right]\right)^{p-1}}.
\]
So by (3.3.25),
\[
X_N=\left[\frac{\lambda
Z}{(1+r)^N}\right]^{\frac{1}{p-1}}=\frac{\lambda^{\frac{1}{p-1}}
Z^{\frac{1}{p-1}}}{(1+r)^{\frac{N}{p-1}}}=\frac{X_0(1+r)^{\frac{Np}{p-1}}}{{\mathbb E}\left[Z^{\frac{p}{p-1}}\right]}\frac{Z^{\frac{1}{p-1}}}{(1+r)^{\frac{N}{p-1}}}=\frac{(1+r)^NX_0Z^{\frac{1}{p-1}}}{{\mathbb E}\left[Z^{\frac{p}{p-1}}\right]}.
\]
\end{proof}


\noindent $\blacktriangleright$ {\bf Exercise 3.8.} The Lagrange Multiplier Theorem used in the solution of Problem 3.3.5 has hypotheses that we did not verify in the solution of that problem. In particular, the theorem states that if the gradient of the constraint function, which in this case is the vector $(p_1\zeta_1,\cdots,p_m\zeta_m)$, is not the zero vector, then the optimal solution must satisfy the Lagrange multiplier equations (3.3.22). This gradient is not the zero vector, so this hypothesis is satisfied. However, even when this hypothesis is satisfied, the theorem does not guarantee that there is an optimal solution; the solution to the Lagrange multiplier equations may in fact minimize the expected utility. The solution could also be neither a maximizer nor a minimizer. Therefore, in this exercise, we outline a different method for verifying that the random variable $X_N$ given by (3.3.25) maximizes the expected utility.

We begin by changing the notation, calling the random variable given by (3.3.25) $X_N^*$ rather than $X_N$. In other words,
\[
X_N^* = I\left(\frac{\lambda}{(1+r)^N}Z\right), \tag{3.6.1}
\]
where $\lambda$ is the solution of equation (3.3.26). This permits us to use the notation $X_N$ for an arbitrary (not necessarily optimal) random variable satisfying (3.3.19). We must show that
\[
{\mathbb E}U(X_N) \le {\mathbb E}U(X_N^*). \tag{3.6.2}
\]

\smallskip

\noindent (i) Fix $y>0$, and show that the function of $x$ given by $U(x)-yx$ is maximized by $x=I(y)$.\footnote{The textbook said ``$y=I(x)$" by mistake.} Conclude that
\[
U(x)-yx \le U(I(y)) - yI(y) \; \mbox{for every $x$.} \tag{3.6.3}
\]

\begin{proof} $\frac{d}{dx}(U(x)-yx)=U'(x)-y$. So
$x=I(y)$ is an extreme point of $U(x)-yx$. Because
$\frac{d^2}{dx^2}(U(x)-yx)=U''(x)\le 0$ ($U$ is concave), $x=I(y)$
is a maximum point. Therefore $U(x)-yx\le U(I(y))-yI(y)$ for every
$x$.
\end{proof}

\noindent (ii) In (3.6.3), replace the dummy variable $x$ by the random variable $X_N$ and replace the dummy variable $y$ by the random variable $\frac{\lambda Z}{(1+r)^N}$. Take expectations of both sides and use (3.3.19) and (3.3.26) to conclude that (3.6.2) holds.

\begin{proof} Following the hint of the problem, we have
\[
{\mathbb E}[U(X_N)]-{\mathbb E}\left[X_N\frac{\lambda Z}{(1+r)^N}\right]
\le {\mathbb E}\left[U\left(I\left(\frac{\lambda Z}{(1+r)^N}\right)\right)\right]-{\mathbb E}\left[\frac{\lambda Z}{(1+r)^N}I\left(\frac{\lambda
Z}{(1+r)^N}\right)\right],
\]
i.e.
\begin{eqnarray*}
& & {\mathbb E}[U(X_N)]-\lambda X_0 \overset{(3.3.19)}{=} {\mathbb E}[U(X_N)]-\lambda \widetilde {\mathbb E}\left[\frac{X_N}{(1+r)^N}\right] = {\mathbb E}[U(X_N)]-{\mathbb E}\left[X_N\frac{\lambda Z}{(1+r)^N}\right] \\
&\le&
{\mathbb E}[U(X_N^*)]-\lambda {\mathbb E}\left[\frac{Z}{(1+r)^N}I\left(\frac{\lambda
Z}{(1+r)^N}\right)\right] \overset{(3.3.26)}{=} {\mathbb E}[U(X_N^*)]-\lambda X_0.
\end{eqnarray*}
 So ${\mathbb E}[U(X_N)]\le {\mathbb E}[U(X_N^*)]$.
\end{proof}

\noindent $\blacktriangleright$ {\bf Exercise 3.9 (Maximizing probability of reaching a goal).} (Kulldorf [30], Heath [19]) A wealthy investor provides a small amount of money $X_0$ for you to use to prove the effectiveness of your investment scheme over the next $N$ periods. You are permitted to invest in the $N$-period binomial model, subject to the condition that the value of your portfolio is never allowed to be negative. If at time $N$ the value of your portfolio $X_N$ is at least $\gamma$, a positive constant specified by the investor, then you will be given a large amount of money to manage for her. Therefore, your problem is the following:

\noindent Maximize
\[
{\mathbb P}(X_N \ge \gamma),
\]
where $X_N$ is generated by a portfolio process beginning with the initial wealth $X_0$ and where the value $X_n$ of your portfolio satisfies
\[
X_n \ge 0, \; n = 1, 2, \cdots, N.
\]

In the way that Problem 3.3.1 was reformulated as Problem 3.3.3, this problem may be reformulated as

\noindent Maximize
\[
{\mathbb P}(X_N \ge \gamma)
\]
subject to
\[
\widetilde {\mathbb E}\frac{X_N}{(1+r)^N} = X_0, \; X_n \ge 0, \; n=1, 2, \cdots, N.
\]

\smallskip

\noindent (i) Show that if $X_N \ge 0$, then $X_n \ge 0$ for all $n$.

\begin{proof} By the martingale property, $X_n=\widetilde {\mathbb E}_n \left[\frac{X_N}{(1+r)^{N-n}}\right]$. So if $X_N\ge 0$, then $X_n\ge 0$ for all $n$.
\end{proof}

\noindent (ii) Consider the function
\[
U(x) = \begin{cases}
0, & \mbox{if $0\le x<\gamma$,} \\
1, & \mbox{if $x\ge \gamma$.}
\end{cases}
\]
Show that for each fixed $y>0$, we have
\[
U(x) - yx \le U(I(y)) - yI(y)\; \forall x \ge 0,
\]
where
\[
I(y) =
\begin{cases}
\gamma, & \mbox{if $0<y \le \frac{1}{\gamma}$,} \\
0, & \mbox{if $y>\frac{1}{\gamma}$.}
\end{cases}
\]
\begin{proof} We consider the following four scenarios.

a) If $0\le x<\gamma$ and $0<y\le
\frac{1}{\gamma}$, then $U(x)-yx=-yx\le 0$ and
$U(I(y))-yI(y)=U(\gamma)-y\gamma=1-y\gamma\ge 0$. So $U(x)-yx\le
U(I(y))-yI(y)$.

b) If $0\le x<\gamma$ and $y>\frac{1}{\gamma}$, then $U(x)-yx=-yx\le
0$ and $U(I(y))-yI(y)=U(0)-y\cdot 0=0$. So $U(x)-yx\le
U(I(y))-yI(y)$.

c) If $x\ge \gamma$ and $0<y\le\frac{1}{\gamma}$, then
$U(x)-yx=1-yx$ and $U(I(y))-yI(y)=U(\gamma)-y\gamma=1-y\gamma\ge
1-yx$. So $U(x)-yx\le U(I(y))-yI(y)$.

d) If $x\ge \gamma$ and $y>\frac{1}{\gamma}$, then $U(x)-yx=1-yx<0$
and $U(I(y))-yI(y)=U(0)-y\cdot 0=0$. So $U(x)-yx\le U(I(y))-yI(y)$.

\end{proof}

\noindent (iii) Assume there is a solution $\lambda$ to the equation
\[
{\mathbb E} \left[\frac{Z}{(1+r)^N} I \left(\frac{\lambda Z}{(1+r)^N}\right)\right] = X_0. \tag{3.6.4}
\]
Following the argument of Exercise 3.8, show that the optimal $X_N$ is given by
\[
X_N^* = I \left(\frac{\lambda Z}{(1+r)^N}\right).
\]
\begin{proof} Using (ii) and set $x=X_N$, $y=\frac{\lambda
Z}{(1+r)^N}$, where $X_N$ is a random variable satisfying
$\widetilde {\mathbb E}\left[\frac{X_N}{(1+r)^N}\right]=X_0$, we have
\[
 {\mathbb E}[U(X_N)]- {\mathbb E}\left[\frac{\lambda Z}{(1+r)^N}X_N\right]
 \le
 {\mathbb E}[U(X_N^*)]- {\mathbb E}\left[\frac{\lambda Z}{(1+r)^N}X_N^*\right].
\]
That is, $ {\mathbb E}[U(X_N)]-\lambda X_0\le  {\mathbb E}[U(X_N^*)]-\lambda X_0$. So
$ {\mathbb E}[U(X_N)]\le  {\mathbb E}[U(X_N^*)]$.
\end{proof}


\noindent (iv) As we did to obtain Problem 3.3.5, let us list the $M=2^N$ possible coin toss sequences, labeling them $\omega^1$, $\cdots$, $\omega^M$, and then define $\zeta_m = \zeta(\omega^{m})$, $p_m = {\mathbb P}(\omega^m)$. However, here we list these sequences in ascending order of $\zeta_m$ i.e., we label the coin toss sequences so that
\[
\zeta_1 \le \zeta_2 \le \cdots \le \zeta_M.
\]
Show that the assumption that there is a solution $\lambda$ to (3.6.4) is equivalent to assuming that for some positive integer $K$ we have $\zeta_K < \zeta_{K+1}$ and
\[
\sum_{m=1}^K \zeta_m p_m = \frac{X_0}{\gamma}. \tag{3.6.5}
\]

\begin{proof} Plug $p_m$ and $\xi_m$ into (3.6.4), we have
\[
X_0=\sum_{m=1}^{2^N}p_m\xi_mI(\lambda
\xi_m)=\sum_{m=1}^{2^N}p_m\xi_m\gamma
1_{\{\lambda\xi_m\le\frac{1}{\gamma}\}}.
\]
So $\frac{X_0}{\gamma}=\sum_{m=1}^{2^N}p_m\xi_m1_{\{\lambda
\xi_m\le\frac{1}{\gamma}\}}$. Suppose there is a solution $\lambda$
to (3.6.4), note $\frac{X_0}{\gamma}>0$, we then can conclude
$\{m:\lambda\xi_m\le \frac{1}{\gamma}\}\ne \emptyset$. Let
$K=\max\{m:\lambda \xi_m\le\frac{1}{\gamma}\}$, then
$\lambda\xi_K\le\frac{1}{\gamma}<\lambda\xi_{K+1}$. So
$\xi_K<\xi_{K+1}$ and $\frac{X_0}{\gamma}=\sum_{m=1}^Kp_m\xi_m$
(Note, however, that $K$ could be $2^N$. In this case, $\xi_{K+1}$
is interpreted as $\infty$. Also, note we are looking for positive
solution $\lambda>0$). Conversely, suppose there exists some $K$ so
that $\xi_K<\xi_{K+1}$ and
$\sum_{m=1}^K\xi_mp_m=\frac{X_0}{\gamma}$. Then we can find
$\lambda>0$, such that $\xi_K<\frac{1}{\lambda \gamma}<\xi_{K+1}$.
For such $\lambda$, we have
\[
{\mathbb E}\left[\frac{Z}{(1+r)^N}I(\frac{\lambda
Z}{(1+r)^N})\right]=\sum_{m=1}^{2^N}p_m\xi_m1_{\{\lambda\xi_m\le\frac{1}{\gamma}\}}\gamma=\sum_{m=1}^Kp_m\xi_m\gamma=X_0.
\]Hence (3.6.4) has a solution.
\end{proof}

\noindent (v) Show that $X_N^*$ is given by
\[
X_N(\omega^m) = \begin{cases}
\gamma, & \mbox{if $m\le K$} \\
0, & \mbox{if $m\ge K+1$}.
\end{cases}
\]

\begin{proof}
\[
X_N^*(\omega^m)=I(\lambda
\xi_m)=\gamma1_{\{\lambda\xi_m\le\frac{1}{\gamma}\}}=
\begin{cases}
\gamma,&\mbox{if $m\le K$,}\\
0, &\mbox{if $m\ge K+1$}.
\end{cases}
\]
\end{proof}


\section{American Derivative Securities}

$\bigstar$ {\bf Comments}:

\medskip

1) Before proceeding to the exercise problems, we first give a brief
summary of pricing American derivative securities as presented in
the textbook. We shall use the notation of the book.

{\it From the buyer's perspective.} At time $n$, if the derivative security
has not been exercised, then the buyer can choose a policy $\tau$
with $\tau\in {\cal S}_n$. The valuation formula for cash flow
(Theorem 2.4.8) gives the fair price of the derivative security
exercised according to $\tau$ (the cash flow sequence $C_n$, $\cdots$, $C_N$ at times $n$, $\cdots$, $N$ are defined by $C_n=1_{\{\tau=n\}}G_n$, $\cdots$, $C_N=1_{\{\tau=N\}}G_N$):
\[
V_n(\tau)=\sum_{k=n}^N\widetilde
{\mathbb E}_n\left[1_{\{\tau=k\}}\frac{G_k}{(1+r)^{k-n}}\right]=\widetilde
{\mathbb E}_n\left[1_{\{\tau\le N\}}\frac{G_{\tau}}{(1+r)^{\tau-n}}\right].
\]
The buyer wants to consider all the possible $\tau$'s, so that he
can find the least upper bound of security value, which will be the
maximum price of the derivative security acceptable to him. This is
the price given by Definition 4.4.1: $V_n=\max_{\tau\in {\cal
S}_n}\widetilde {\mathbb E}_n \left[1_{\{\tau\le
N\}}\frac{1}{(1+r)^{\tau-n}}G_{\tau} \right]$.

{\it From the seller's perspective.} A price process $(V_n)_{0\le n\le N}$
is acceptable to him if and only if at each time $n$, (i) $V_n\ge G_n$ (selling price is sufficient to cover potential obligation) and (ii) he needs
no further investing into the portfolio as time goes by (no follow-up cost to maintain (i)).

Formally, the seller can find hedging process $(\Delta_n)_{0\le n\le N}$ and cash flow process $(C_n)_{0\le n\le
N}$ such that $C_n\ge 0$ and $V_{n+1}=\Delta_nS_{n+1}+(1+r)(V_n-C_n-\Delta_nS_n) \ge G_{n+1}$. Since
$\left\{\frac{S_n}{(1+r)^n}\right\}_{0\le n\le N}$ is a martingale under the
risk-neutral measure $\widetilde {\mathbb P}$, we conclude
\[
\widetilde
{\mathbb E}_n\left[\frac{V_{n+1}}{(1+r)^{n+1}}\right]-\frac{V_n}{(1+r)^n}=-\frac{C_n}{(1+r)^n}\le
0,
\]i.e. $\left\{\frac{V_n}{(1+r)^n}\right\}_{0\le n\le N}$ is a supermartingale under $\widetilde {\mathbb P}$. This inspires us to check if the converse is also true. This is
exactly the content of Theorem 4.4.4. So $(V_n)_{0\le n\le N}$ is
the value process of a portfolio that needs no further investing if
and only if $\left\{\frac{V_n}{(1+r)^n}\right\}_{0\le n\le N}$ is a
supermartingale under $\widetilde {\mathbb P}$ (note this is independent of
the requirement $V_n\ge G_n$). In summary, a price process
$(V_n)_{0\le n\le N}$ is acceptable to the seller if and only if (i)
$V_n\ge G_n$; (ii) $\left\{\frac{V_n}{(1+r)^n}\right\}_{0\le n\le N}$
is a supermartingale under $\widetilde {\mathbb P}$.

Theorem 4.4.2 shows the buyer's upper bound is the seller's lower
bound. So it gives the price acceptable to both. Theorem 4.4.3 gives
a specific algorithm for calculating the price, Theorem 4.4.4
establishes the one-to-one correspondence between super-replication
and supermartingale property, and finally, Theorem 4.4.5 shows how
to decide on the optimal exercise policy.

\medskip

2) The definition of a stopping time typically seen in textbooks is
\[
\{\tau = n \} \in {\cal F}_n := \sigma(\omega_1, \cdots, \omega_n), \; n = 0, 1, 2, \cdots, N.
\]
To see Definition 4.3.1 agrees with this version of definition, note a typical element of ${\cal F}_n$ has the form
$\{\omega: \omega_1 \in A_1, \cdots, \omega_n \in A_n \}$. So $\{\tau=n\}$ imposes conditions on $\omega_1$, $\omega_2$, $\cdots$, $\omega_n$ only.

\medskip

3) Comment on Theorem 4.4.4 (Replication of path-dependent American derivatives): in the proof, the condition that $(V_n)_{n=0}^N$ comes from Definition 4.4.1 is only used for $V_n \ge G_n$. The proof applies to any $\widetilde {\mathbb P}$-supermartingale $\left\{\frac{V_n}{(1+r)^n}\right\}_n$. This justifies the paragraph immediately after Theorem 4.4.2. Conversely, if $(V_n)_n$ can be replicated via formula (4.4.16) with $C_n\ge 0$, then $\left\{\frac{V_n}{(1+r)^n}\right\}_n$ is a $\widetilde {\mathbb P}$-supermartingale. That is, if we can replicate a stochastic process $(V_n)_n$ by money market account and stock without injecting more cash into the portfolio, then $\left\{\frac{V_n}{(1+r)^n}\right\}_n$ is a $\widetilde {\mathbb P}$-supermartingale.


\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 4.1.} In the three-period model of Figure 1.2.2 of Chapter 1, let the interest rate be $r=\frac{1}{4}$ so the risk-neutral probabilities are $\tilde p = \tilde q = \frac{1}{2}$.

\smallskip

\noindent (i) Determine the price at time zero, denoted $V_0^P$, of the American put that expires at time three and has intrinsic value $g_P(s)=(4-s)^+$.

\begin{solution}
At time three, the payoff of the put is
\begin{eqnarray*}
&& V_3^P(HHH) = (4-32)^+ = 0 \\
&& V_3^P(HHT) = V_3^P(HTH) = V_3^P(THH) = (4-8)^+ = 0 \\
&& V_3^P(HTT) = V_3^P(THT) = V_3^P(TTH) = (4-2)^+ = 2 \\
&& V_3^P(TTT) = (4-0.5)^+ = 3.5.
\end{eqnarray*}
At time two, the value of the put is calculated by
\begin{eqnarray*}
V_2^P(\cdot) = \max\left\{(4-S_2)^+, \widetilde {\mathbb E}_2\left[\frac{V_3^P}{1+r}\right]\right\}
= \max\left\{(4-S_2)^+,\frac{\tilde p V_3^P(\cdot H) + \tilde q V_3^P(\cdot T)}{1+r}\right\}.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
&& V_2^P(HH) =  \max \left\{[4-S_2(HH)]^+, \frac{2}{5}[V_3^P(HHH)+V_3^P(HHT)] \right\} = \max \left\{(4-16)^+, \frac{2}{5}(0+0)\right\} = 0 \\
&& V_2^P(HT) = \max \left\{[4-S_2(HT)]^+, \frac{2}{5}[V_3^P(HTH)+V_3^P(HTT)] \right\} = \max \left\{(4-4)^+, \frac{2}{5}(0+2)\right\} = 0.8 \\
&& V_2^P(TH) =  \max \left\{[4-S_2(TH)]^+, \frac{2}{5}[V_3^P(THH)+V_3^P(THT)] \right\} = \max \left\{(4-4)^+, \frac{2}{5}(0+2)\right\} = 0.8 \\
&& V_2^P(TT) = \max \left\{[4-S_2(TT)]^+, \frac{2}{5}[V_3^P(TTH)+V_3^P(TTT)] \right\} = \max \left\{(4-1)^+, \frac{2}{5}(2+3.5)\right\} = 3 \\
\end{eqnarray*}
A time one, the value of the put is calculated by
\[
V_1^P(\cdot) = \max\left\{(4-S_1)^+, \widetilde {\mathbb E}_1\left[\frac{V_2^P}{1+r}\right]\right\}
= \max\left\{(4-S_1)^+,\frac{\tilde p V_2^P(\cdot H) + \tilde q V_2^P(\cdot T)}{1+r}\right\}
\]
Therefore
\begin{eqnarray*}
&& V_1^P(H) = \max \left\{[4-S_1(H)]^+, \frac{2}{5}[V_2^P(HH)+V_2^P(HT)] \right\} = \max \left\{(4-8)^+, \frac{2}{5}(0+0.8) \right\} = 0.32 \\
&& V_1^P(T) = \max \left\{[4-S_1(T)]^+, \frac{2}{5}[V_2^P(TH)+V_2^P(TT)] \right\} = \max \left\{(4-2)^+, \frac{2}{5}(0.8+3) \right\} = 2.
\end{eqnarray*}
Finally, the value of the put at time zero is
\[
V_0^P = \max \left\{(4-S_0)^+, \frac{2}{5}[V_1^P(H)+V_1^P(T)] \right\} = \max \left\{(4-4)^+, \frac{2}{5}(0.32+2) \right\} = 0.928.
\]
\end{solution}

\noindent (ii) Determine the price at time zero, denoted $V_0^C$, of the American call that expires at time three and has intrinsic value $g_C(s) = (s-4)^+$.

\begin{solution}By Theorem 4.5.1, at the time zero, the price of the American call is the same as the price of the European call with the same strike. Therefore, at time three,
\begin{eqnarray*}
&& V_3^C(HHH) = (32-4)^+ = 28, \\
&& V_3^C(HHT) = V_3^C(HTH) = V_3^C(THH) = (8-4)^+ = 4, \\
&& V_3^C(HTT) = V_3^C(THT) = V_3^C(TTH) = (2-4)^+ = 0, \\
&& V_3^C(TTT) = (0.5-4)^+ = 0. \\
\end{eqnarray*}
At time two,
\begin{eqnarray*}
V_2^C(HH) = \frac{2}{5}(28+4)= 12.8, \;
   V_2^C(HT) = V_2^C(TH) = \frac{2}{5}(4+0) = 1.6, \;
   V_2^C(TT) = \frac{2}{5}(0+0) = 0. \\
\end{eqnarray*}
At time one,
\begin{eqnarray*}
V_1^C(H) = \frac{2}{5}(12.8+1.6) = 5.76, \;
V_1^C(T) = \frac{2}{5}(1.6+0) = 0.64.
\end{eqnarray*}
And finally, at time zero,
\[
V_0^C = \frac{2}{5}(5.76+0.64) = 2.56.
\]
\end{solution}

\noindent (iii) Determine the price at time zero, denoted $V_0^S$, of the American straddle that expires at time three and has intrinsic value $g_S(s) = g_P(s) + g_C(s)$.

\begin{solution}$g_S(s)=|4-s|$. At time three,
\begin{eqnarray*}
& & V_3^S(HHH) = |4-32| = 28, \\
& & V_3^S(HHT)=V_3^S(HTH)=V_3^S(THH) = |4-8|=4, \\
& & V_3^S(HTT) = V_3^S(THT) = V_3^S(TTH) = |4-2|=2, \\
& & V_3^S(TTT) = |4-0.5| = 3.5.
\end{eqnarray*}
At time two,
\begin{eqnarray*}
& & V_2^S(HH) = \max\left\{|4-16|, \frac{2}{5}(28+4) \right\} = 12.8,\\
& & V_2^S(HT) = V_2^S(TH) = \max\left\{|4-4|,\frac{2}{5}(4+2)\right\} = 2.4,\\
& & V_2^S(TT) = \max\left\{|4-1|,\frac{2}{5}(2+0.5)\right\} = 3.
\end{eqnarray*}
At time one,
\begin{eqnarray*}
& & V_1^S(H) = \max \left\{|4-8|, \frac{2}{5}(12.8+2.4)\right\} = 6.08, \\
& & V_1^S(T) = \max \left\{|4-2|, \frac{2}{5}(2.4+3)\right\} = 2.16.
\end{eqnarray*}
Finally, at time zero,
\[
V_0^S = \max\left\{|4-4|, \frac{2}{5}(6.08+2.16)\right\} = 3.296.
\]
\end{solution}

\noindent (iv) Explain why $V_0^S < V_0^P + V_0^C$.

\begin{solution} In our previous calculations,
\[
V_0^S = 3.296 < V_0^P + V_0^C = 0.928 + 2.56.
\]

To explain this, we first note the simple inequality
\[
\max(a_1,b_1)+\max(a_2,b_2)\ge \max(a_1+a_2,b_1+b_2),
\]
where ``$>$" holds if and only if $b_1>a_1$ and $b_2<a_2$ {\it or} $b_1<a_1$ and
$b_2>a_2$.

At time $N$, $V_N^S = g_S(S_N) = g_P(S_N) + g_C(S_N) = V_N^P + V_N^C$.  By induction, we can show
\begin{eqnarray*}
V_n^S &=&\max\left\{g_S(S_n),\frac{\tilde pV^S_{n+1}+\tilde
qV^S_{n+1}}{1+r}\right\}\\
&\le&\max\left\{g_P(S_n)+g_C(S_n),\frac{\tilde
pV^P_{n+1}+\tilde q V^P_{n+1}}{1+r}+\frac{\tilde
pV^C_{n+1}+\tilde q
V^C_{n+1}}{1+r}\right\}\\
&\le&\max\left\{g_P(S_n),\frac{\tilde pV^P_{n+1}+\tilde
qV^P_{n+1}}{1+r}\right\}+\max\left\{g_C(S_n),\frac{\tilde
pV^C_{n+1}+\tilde qV^C_{n+1}}{1+r}\right\}\\
&=&V_n^P+V_n^C,
\end{eqnarray*}
$n = 0, 1, \cdots, N-1$.

To see when ``$<$" holds, note for American call options, we always have $g_C(S_n) < \frac{\tilde
pV^C_{n+1}+\tilde qV^C_{n+1}}{1+r}$ for $n = 0, 1, \cdots, N-1$. In view of our observation of the simple inequality at the beginning of our solution, ``$<$" holds whenever
\[
g_P(S_n) > \frac{\tilde
pV^P_{n+1}+\tilde q V^P_{n+1}}{1+r}.
\]
This is typically the case at the optimal exercise time of the put.

Beside the mathematical argument, intuitively, a straddle is not an American put plus an American call, since when exercised, a straddle has a payoff equal to the sum of the payoff of a put and the payoff of a call. However, the exercise time of a put typically differs from that of a call. So intuitively, we must have $V_0^S < V_0^P + V_0^C$.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.2.} In Example 4.2.1, we computed the time-zero value of the American put with strike price $5$ to be $1.36$. Consider an agent who borrows 1.36 at time zero and buys the put. Explain how this agent can generate sufficient funds to pay off his loan (which grows by 25\% each period) by trading in the stock and money markets and optimally exercising the put.

\begin{solution}For this problem, we need Figure 4.2.1,
Figure 4.4.1 and Figure 4.4.2. Then
\[
\Delta_1(H)=\frac{V_2(HH)-V_2(HT)}{S_2(HH)-S_2(HT)}=-\frac{1}{12},\;\Delta_1(T)=\frac{V_2(TH)-V_2(TT)}{S_2(TH)-S_2(TT)}=-1,
\]and
\[
\Delta_0=\frac{V_1(H)-V_1(T)}{S_1(H)-S_1(T)}\approx -0.433.
\]The optimal exercise time is $\tau=\inf\{n:V_n=G_n\}$. So
\[
\tau(HH)=\infty,\;\tau(HT)=2,\;\tau(TH)=\tau(TT)=1.
\]
Therefore, the agent borrows 1.36 at time zero and buys the put. At
the same time, to hedge the long position, he needs to borrow again
and buy 0.433 shares of stock at time zero.

At time one, if the result of coin toss is tail and the stock price
goes down to 2, the value of the portfolio is
$X_1(T)=(1+r)(-1.36-0.433S_0)+0.433S_1(T)=(1+\frac{1}{4})(-1.36-0.433\times
4)+0.433\times 2=-3$. The agent should exercise the put at time one
and get 3 to pay off his debt.

At time one, if the result of coin toss is head and the stock price
goes up to 8, the value of the portfolio is
$X_1(H)=(1+r)(-1.36-0.433S_0)+0.433S_1(H)=-0.4$. The agent should
borrow to buy $\frac{1}{12}$ shares of stock. At time two, if the
result of coin toss is head and the stock price goes up to 16, the
value of the portfolio is
$X_2(HH)=(1+r)(X_1(H)-\frac{1}{12}S_1(H))+\frac{1}{12}S_2(HH)=0$,
and the agent should let the put expire. If at time two, the result
of coin toss is tail and the stock price goes down to 4, the value
of the portfolio is
$X_2(HT)=(1+r)(X_1(H)-\frac{1}{12}S_1(H))+\frac{1}{12}S_2(HT)=-1$.
The agent should exercise the put to get 1. This will pay off his
debt.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.3.} In the three-period model of Figure 1.2.2 of Chapter 1, let the interest rate be $r=\frac{1}{4}$ so the risk-neutral probabilities are $\tilde p = \tilde q = \frac{1}{2}$. Find the time-zero price and optimal exercise policy (optimal stopping time) for the path-dependent American derivative security whose intrinsic value at each time $n$, $n=0, 1, 2, 3$, is $\left(4-\frac{1}{n+1}\sum_{j=0}^n S_j\right)^+$. This intrinsic value is a put on the average stock price between time zero and time $n$.

\begin{solution} We need Figure 1.2.2 for this problem,
and calculate the intrinsic value process and price process of the
put as follows.

For the intrinsic value process, $G_0=0$, $G_1(T)=1$,
$G_2(TH)=\frac{2}{3}$, $G_2(TT)=\frac{5}{3}$, $G_3(THT)=1$,
$G_3(TTH)=1.75$, $G_3(TTT)=2.125$. All the other outcomes of $G$ is
negative.

For the price process, $V_0=0.4$, $V_1(T)=1$, $V_1(TH)=\frac{2}{3}$,
$V_1(TT)=\frac{5}{3}$, $V_3(THT)=1$, $V_3(TTH)=1.75$,
$V_3(TTT)=2.125$. All the other outcomes of $V$ is zero.

Therefore the time-zero price of the derivative security is 0.4 and
the optimal exercise time satisfies
\[
\tau(\omega)=\left\{
\begin{matrix}
\infty & \mbox{if $\omega_1=H$,}\\
1& \mbox{if $\omega_1=T$.}
\end{matrix}
\right.
\]
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.4.} Consider the American put of Example 4.2.1, which
has strike price 5. Suppose at time zero we sell this put to a purchaser who has inside information about the stock movements and uses the exercise rule $\rho$ of (4.3.2). In particular, if the first toss is going to result in $H$, the owner of the put exercises at time zero, when the put has intrinsic value 1. If the first toss results in $T$ and the second toss is going to result in $H$, the owner exercises at time one, when the put has intrinsic value 3. If the first two tosses result in $TT$, the owner exercises at time two, when the intrinsic value is 4. In summary, the owner of the put has the payoff random variable
\[
Y(HH)=1, \; Y(HT) = 1, \; Y(TH) = 3, \; Y(TT) = 4. \tag{4.8.1}
\]
The risk-neutral expected value of this payoff, discounted from the time of payment back to zero, is
\[
\widetilde {\mathbb E} \left[\left(\frac{4}{5}\right)^{\rho}Y\right] = \frac{1}{4} \left[1+1+\frac{4}{5}\cdot 3 + \frac{16}{25}\cdot 4\right] = 1.74. \tag{4.8.2}
\]
The time-zero price of the put computed in Example 4.2.1 is only 1.36. Do we need to charge the insider more than this amount if we are going to successfully hedge our short position after selling the put to her? Explain why or why not.

\begin{solution}$1.36$ is the cost of super-replicating
the American derivative security. It enables us to construct a
portfolio sufficient to pay off the derivative security, no matter
when the derivative security is exercised. So to hedge our short
position after selling the put, there is no need to charge the
insider more than $1.36$.

To understand the paradox that the risk-neutral expected value 1.74 of the insider's payoff is greater than the fair price 1.36, we note the rationale for risk-neutral pricing is the existence of a self-financing replicating portfolio:
\[
X_{n+1} = \Delta_n S_{n+1} + (1+r)(X_n - C_n - \Delta_n S_n), \; n = 0, 1, \cdots, N-1,
\]
where the hedging process $(\Delta_n)_n$ is assumed to be adapted, i.e. $\Delta_n \in {\cal F}_n$. This is in contradiction of the use of inside information. Therefore, it is fishy to use risk-neutral pricing formula when insider information exists.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.5.} In equation (4.4.5), the maximum is computed over all stopping times in ${\cal S}_0$. List all the stopping times in ${\cal S}_0$ (there are 26), and from among them, list the stopping times that never exercise when the option is out of the money (there are 11). For each stopping time $\tau$ in the latter set, compute $\widetilde {\mathbb E}\left[1_{\{\tau \le 2\}} \left(\frac{4}{5}\right)^{\tau} G_{\tau}\right]$. Verify that the largest value for this quantity is given by the stopping time of (4.4.6), the one that makes this quantity equal to the 1.36 computed in (4.4.7).

\begin{solution} The stopping times in ${\cal S}_0$ are
\newline (1) $\tau\equiv 0$;
\newline (2) $\tau\equiv 1$;
\newline (3) $\tau(HT)=\tau(HH)=1$, $\tau(TH)$, $\tau(TT)\in
\{2,\infty\}$ (4 different ones);
\newline (4) $\tau(HT)$, $\tau(HH)\in \{2,\infty\}$,
$\tau(TH)=\tau(TT)=1$ (4 different ones);
\newline (5) $\tau(HT),\tau(HH),\tau(TH),\tau(TT)\in \{2,\infty\}$
(16 different ones).

When the option is out of money, the following stopping times do not
exercise
\newline (i) $\tau\equiv 0$;
\newline (ii) $\tau(HT)\in \{2,\infty\}$, $\tau(HH)=\infty$,
$\tau(TH),\tau(TT)\in \{2,\infty\}$ (8 different ones);
\newline (iii) $\tau(HT)\in \{2,\infty\}$, $\tau(HH)=\infty$,
$\tau(TH)=\tau(TT)=1$ (2 different ones).

For (i), $\widetilde {\mathbb E}\left[1_{\{\tau\le
2\}}\left(\frac{4}{5}\right)^{\tau}G_{\tau}\right]=G_0=1$. For (ii), $\widetilde
{\mathbb E}\left[1_{\{\tau\le 2\}}\left(\frac{4}{5}\right)^{\tau}G_{\tau}\right] \le \widetilde
{\mathbb E}\left[1_{\{\tau^*\le 2\}}\left(\frac{4}{5}\right)^{\tau^*}G_{\tau^*}\right]$, where
$\tau^*(HT)=2$, $\tau^*(HH)=\infty$, $\tau^*(TH)=\tau^*(TT)=2$.  So
\[
\widetilde {\mathbb E}\left[1_{\{\tau^*\le
2\}}\left(\frac{4}{5}\right)^{\tau^*}G_{\tau^*}\right]=\frac{1}{4}\left[\left(\frac{4}{5}\right)^2\cdot
1 + \left(\frac{4}{5}\right)^2 (1+4)\right]=0.96.
\]
For (iii), $\widetilde
{\mathbb E}\left[1_{\{\tau\le 2\}}\left(\frac{4}{5}\right)^{\tau}G_{\tau}\right]$ has the biggest
value when $\tau$ satisfies $\tau(HT)=2$, $\tau(HH)=\infty$,
$\tau(TH)=\tau(TT)=1$. This value is 1.36.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.6 (Estimating American put prices).} For each $n$, where $n=0, 1, \cdots, N$, let $G_n$ be a random variable depending on the first $n$ coin tosses. The time-zero value of a derivative security that can be exercised at nay time $n\le N$ for payoff $G_n$ but {\it must be exercised at time $N$ if it has not been exercised before that time} is
\[
V_0 = \max_{\tau \in {\cal S}_0, \tau \le N} \widetilde {\mathbb E} \left[\frac{1}{(1+r)^{\tau}}G_{\tau}\right]. \tag{4.8.3}
\]
In contrast to equation (4.4.1) in the Definition 4.4.1 for American derivative securities, here we consider only stopping times that take one of the values $0, 1, \cdots, N$ and not the value $\infty$.

\smallskip

\noindent (i) Consider $G_n = K - S_n$, the derivative security that permits its owner to sell one share of stock for payment $K$ at any time up to and including $N$, but if the owner does not sell by time $N$, then she must do so at time $N$. Show that the optimal exercise policy is to sell the stock at time zero and that the value of this derivative security is $K-S_0$.

\begin{proof} The value of the put at time $N$,
if it is not exercised at previous times, is $K-S_N$. Hence
\[
V_{N-1}=\max \left\{K-S_{N-1},\widetilde
{\mathbb E}_{N-1}\left[\frac{V_N}{1+r}\right]\right\}=\max\left\{K-S_{N-1},\frac{K}{1+r}-S_{N-1}\right\}=K-S_{N-1}.
\]
The second equality comes from the fact that discounted stock price
process is a martingale under risk-neutral probability. By
induction, we can show $V_n=K-S_n$ $(0\le n\le N)$. So by Theorem
4.4.5, the optimal exercise policy is to sell the stock at time zero
and the value of this derivative security is $K-S_0$.
\end{proof}

\begin{remark} We cheated a little bit by using American algorithm
and Theorem 4.4.5, since they are developed for the case where
$\tau$ is allowed to be $\infty$. But intuitively, results in this
chapter should still hold for the case $\tau\le N$, provided we
replace ``$\max\{G_n,0\}$" with ``$G_n$".
\end{remark}


\noindent (ii) Explain why a portfolio that holds the derivative security in (i) and a European call with strike $K$ and expiration time $N$ is at least as valuable as an American put struck at $K$ with expiration time $N$. Denote the time-zero value of the European call by $V_0^{EC}$ and the time-zero value of the American put by $V_0^{AP}$. Conclude that the upper bound
\[
V_0^{AP} \le K - S_0 + V_0^{EC} \tag{4.8.4}
\]
on $V_0^{AP}$ holds.

\begin{proof} This is because at time $N$, if we have to
exercise the put and $K-S_N<0$, we can exercise the European call to
set off the negative payoff. In effect, throughout the portfolio's
lifetime, the portfolio has intrinsic values no less than that of an
American put stuck at $K$ with expiration time $N$. So, we must have
$V_0^{AP}\le K-S_0+V_0^{EC}$.
\end{proof}

\noindent (iii) Use put-call parity (Exercise 2.11 of Chapter 2) to derive the lower bound on $V_0^{AP}$:
\[
\frac{K}{(1+r)^N} - S_0 + V_0^{EC} \le V_0^{AP}. \tag{4.8.5}
\]

\begin{proof} Let $V_0^{EP}$ denote the time-zero value of a
European put with strike $K$ and expiration time $N$. Then
\[
V_0^{AP}\ge V_0^{EP}=V_0^{EC}-\widetilde
{\mathbb E} \left[\frac{S_N-K}{(1+r)^N} \right]=V_0^{EC}-S_0+\frac{K}{(1+r)^N}.
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 4.7.} For the class of derivative securities described in Exercise 4.6 whose time-zero price is given by (4.8.3), let $G_n = S_n - K$. This derivative security permits its owner to buy one share of stock in exchange for a payment of $K$ at any time up to the expiration time $N$. If the purchase has not been made at time $N$, it must be made then. Determine the time-zero value and optimal exercise policy for this derivative security.\footnote{Compare with American call option, Theorem 4.5.1.}

\begin{solution} $V_N=S_N-K$,
\[
V_{N-1}=\max \left\{S_{N-1}-K,\widetilde
{\mathbb E}_{N-1}\left[\frac{V_N}{1+r}\right] \right\}=\max \left\{S_{N-1}-K,S_{N-1}-\frac{K}{1+r} \right\}=S_{N-1}-\frac{K}{1+r}.
\]
By induction, we can prove $V_n=S_n-\frac{K}{(1+r)^{N-n}}$ $(0\le
n\le N)$ and $V_n>G_n$ for $0\le n\le N-1$. So the time-zero value
is $S_0-\frac{K}{(1+r)^N}$ and the optimal exercise time is $N$.
\end{solution}


\section{Random Walk}


\noindent $\blacktriangleright$ {\bf Exercise 5.1.} For the symmetric random walk, consider the first passage time $\tau_m$ to the level $m$. The random variable $\tau_2-\tau_1$ is the number of steps required for the random walk to rise from level 1 to level 2, and this random variable has the same distribution as $\tau_1$, the number of steps required for the random walk to rise from level 0 to level 1. Furthermore, $\tau_2-\tau_1$ and $\tau_1$ are independent of one another; the latter depends only on the coin toss $1, 2, \cdots, \tau_1$, and the former depends only on the coin tosses $\tau_1+1$, $\tau_1+2$, $\cdots$, $\tau_2$.

\smallskip

\noindent (i) Use these facts to explain why
\[
{\mathbb E}\alpha^{\tau_2} = ({\mathbb E}\alpha^{\tau_1})^2 \; \mbox{for all $\alpha \in (0,1)$.}
\]

\begin{proof}
${\mathbb E}[\alpha^{\tau_2}]={\mathbb E}[\alpha^{(\tau_2-\tau_1)+\tau_1}]={\mathbb E}[\alpha^{(\tau_2-\tau_1)}]{\mathbb E}[\alpha^{\tau_1}]={\mathbb E}[\alpha^{\tau_1}]^2$.
\end{proof}

\begin{remark}
The i.i.d. property of $\tau_2-\tau_1$ and $\tau_1$ can be proven by strong Markov property of random walk.
\end{remark}

\noindent (ii) Without using (5.2.13), explain why for any positive integer $m$ we must have
\[
{\mathbb E} \alpha^{\tau_m} = ({\mathbb E}\alpha^{\tau_1})^m \; \mbox{for all $\alpha \in (0,1)$.} \tag{5.7.1}
\]

\begin{proof} If we define $M_n^{(m)}=M_{n+\tau_m}-M_{\tau_m}$
$(m=1,2,\cdots)$, then $(M_{\cdot}^{(m)})_m$ as random functions are
i.i.d. with distributions the same as that of $M$. So
$\tau_{m+1}-\tau_m=\inf\{n:M_n^{(m)}=1\}$ are i.i.d. with
distributions the same as that of $\tau_1$. Therefore
\[
{\mathbb E}[\alpha^{\tau_m}]={\mathbb E}[\alpha^{(\tau_m-\tau_{m-1})+(\tau_{m-1}-\tau_{m-2})+\cdots+\tau_1}]={\mathbb E}[\alpha^{\tau_1}]^m.
\]
\end{proof}

\begin{remark}
For those concerned with the meaning of ``$(M_{\cdot}^{(m)})_m$ as random functions...", note the function space $E:= \{f: {\mathbb N} \to {\mathbb Z}\}$ is a Polish space with metric $|\!|f-g|\!|=\sum_{n=1}^{\infty} \frac{|f_n-g_n|}{2^n}$. Then each $M_{\cdot}^{(m)}:\Omega \to E$ is a random element from one probability space to another measurable space so that we can talk about its distribution.
\end{remark}

\noindent (iii) Would equation (5.7.1) still hold if the random walk is not symmetric? Explain why or why not.

\begin{solution}Yes, it would still hold since the strong Markov property of random walk, hence the i.i.d. property of $(\tau_{m+1}-\tau_m)_{m=0}^{\infty}$, still holds and the argument of (ii) still works for asymmetric random walk.
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.2 (First passage time for random walk with upward drift).} Consider the asymmetric random walk with probability $p$ for an up step and probability $q=1-p$ for a down step, where $\frac{1}{2} < p < 1$ so that $0<q<\frac{1}{2}$. In the notation of (5.2.1), let $\tau_1$ be the first time the random walk starting from level $0$ reaches level 1. If the random walk never reaches this level, then $\tau_1=\infty$.

\smallskip

\noindent (i) Define $f(\sigma)=pe^{\sigma}+qe^{-\sigma}$. Show that $f(\sigma)>1$ for all $\sigma>0$.

\begin{proof}
For all $\sigma>0$, $f'(\sigma)=pe^{\sigma}-qe^{-\sigma} > p - q > 0$. So $f(\sigma)>f(0)=1$ for all $\sigma>0$.
\end{proof}

\noindent (ii) Show that, when $\sigma>0$, the process
\[
S_n = e^{\sigma M_n} \left(\frac{1}{f(\sigma)}\right)^n
\]
is a martingale.

\begin{proof} ${\mathbb E}_n \left[\frac{S_{n+1}}{S_n} \right]={\mathbb E}_n \left[e^{\sigma
X_{n+1}}\frac{1}{f(\sigma)}\right]=pe^{\sigma}\frac{1}{f(\sigma)}+qe^{-\sigma}\frac{1}{f(\sigma)}=1$.
\end{proof}

\noindent (iii) Show that, for $\sigma>0$,
\[
e^{-\sigma} = {\mathbb E} \left[1_{\{\tau_1<\infty\}} \left(\frac{1}{f(\sigma)}\right)^{\tau_1} \right].
\]
Conclude that ${\mathbb P}(\tau_1 < \infty)=1$.

\begin{proof} By optional sampling theorem (Theorem 4.3.2),
${\mathbb E}[S_{n\wedge\tau_1}]={\mathbb E}[S_0]=1$. Note
\[
S_{n\wedge\tau_1}=e^{\sigma
M_{n\wedge\tau_1}} \left(\frac{1}{f(\sigma)} \right)^{n\wedge\tau_1}\le
e^{\sigma}
\]
for all $\sigma>0$, by bounded convergence theorem and the fact $0 < 1_{\{\tau_1=\infty\}}S_{n\wedge\tau_1} \le e^{\sigma}\left(\frac{1}{f(\sigma)}\right)^n \to 0$ as $n\to\infty$, we have
\[
{\mathbb E}[1_{\{\tau_1<\infty\}}S_{\tau_1}]={\mathbb E}[\lim_{n\to\infty}S_{n\wedge\tau_1}]=\lim_{n\to\infty}{\mathbb E}[S_{n\wedge\tau_1}]=1.
\]
That is,
${\mathbb E}\left[1_{\{\tau_1<\infty\}}e^{\sigma}\left(\frac{1}{f(\sigma)}\right)^{\tau_1}\right]=1$.
So
$e^{-\sigma}={\mathbb E}\left[1_{\{\tau_1<\infty\}}\left(\frac{1}{f(\sigma)}\right)^{\tau_1}\right]$.
Let $\sigma\downarrow 0$, again by bounded convergence theorem,
$1={\mathbb E}\left[1_{\{\tau_1<\infty\}}\left(\frac{1}{f(0)}\right)^{\tau_1}\right]=P(\tau_1<\infty)$.
\end{proof}

\noindent (iv) Compute ${\mathbb E}\alpha^{\tau_1}$ for $\alpha \in (0,1)$.

\begin{solution} Set
$\alpha=\frac{1}{f(\sigma)}=\frac{1}{pe^{\sigma}+qe^{-\sigma}}$,
then as $\sigma$ varies from $0$ to $\infty$, $\alpha$ can take all
the values in $(0,1)$. Write $\sigma$ in terms of $\alpha$, we have
(note $4pq\alpha^2<4(\frac{p+q}{2})^2\cdot 1^2=1$)
\[
e^{\sigma}=\frac{1\pm\sqrt{1-4pq\alpha^2}}{2p\alpha}.
\]
We want to choose
$\sigma>0$, so we should take
$\sigma=\ln\left(\frac{1+\sqrt{1-4pq\alpha^2}}{2p\alpha}\right)$. Therefore
\[
{\mathbb E}[\alpha^{\tau_1}]=\frac{2p\alpha}{1+\sqrt{1-4pq\alpha^2}}=\frac{1-\sqrt{1-4pq\alpha^2}}{2q\alpha}.
\]
\end{solution}

\noindent (v) Compute ${\mathbb E}\tau_1$.

\begin{solution}
$\frac{\partial}{\partial
\alpha} {\mathbb E}[\alpha^{\tau_1}]= {\mathbb E}[\frac{\partial}{\partial
\alpha}\alpha^{\tau_1}]= {\mathbb E}[\tau_1\alpha^{\tau_1-1}]$, and
\begin{eqnarray*}
& &\left(\frac{1-\sqrt{1-4pq\alpha^2}}{2q\alpha}\right)'\\
&=&\frac{1}{2q}\left[(1-\sqrt{1-4pq\alpha^2})\alpha^{-1}\right]'\\
&=&\frac{1}{2q}\left[-\frac{1}{2}(1-4pq\alpha^2)^{-\frac{1}{2}}(-4pq2\alpha)\alpha^{-1}+(1-\sqrt{1-4pq\alpha^2})(-1)\alpha^2\right].
\end{eqnarray*}
Therefore
\[
{\mathbb E}[\tau_1]=\lim_{\alpha\uparrow
1}\frac{\partial}{\partial\alpha}{\mathbb E}[\alpha^{\tau_1}]=\frac{1}{2q}\left[-\frac{1}{2}(1-4pq)^{-\frac{1}{2}}(-8pq)-(1-\sqrt{1-4pq})\right]=\frac{1}{2p-1}=\frac{1}{p-q}.
\]
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.3 (First passage time for random walk with downward drift).} Modify Exercise 5.2 by assuming $0<p<\frac{1}{2}$ so that $\frac{1}{2} < q < 1$.

\smallskip

\noindent (i) Find a positive number $\sigma_0$ such that the function $f(\sigma)=pe^{\sigma}+qe^{-\sigma}$ satisfies $f(\sigma_0)=1$ and $f(\sigma)>1$ for all $\sigma>\sigma_0$.

\begin{solution}Solve the equation
$pe^{\sigma}+qe^{-\sigma}=1$ and a positive solution is
\[
\ln\frac{1+\sqrt{1-4pq}}{2p}=\ln\frac{1-p}{p}=\ln q-\ln p.
\]
Set
$\sigma_0=\ln q-\ln p$, then $f(\sigma_0)=1$ and
\[
f'(\sigma) = pe^{\sigma}-qe^{-\sigma} = q e^{-\sigma} \left[e^{2\left(\sigma-\frac{\ln q - \ln p}{2}\right)}-1\right] > 0
\]
for $\sigma>\sigma_0$. So $f(\sigma)> f(\sigma_0) = 1$ for all $\sigma>\sigma_0$.
\end{solution}

\noindent (ii) Determine ${\mathbb P}\{\tau_1<\infty\}$. (This quantity is no longer equal to 1.)

\begin{solution} As in Exercise 5.2, $S_n=e^{\sigma
M_n} \left(\frac{1}{f(\sigma)} \right)^n$ is a martingale, and
\[
1={\mathbb E}[S_0]=E[S_{n\wedge\tau_1}]={\mathbb E}\left[e^{\sigma
M_{n\wedge\tau_1}}\left(\frac{1}{f(\sigma)}\right)^{\tau_1\wedge n}\right].
\]
Suppose
$\sigma>\sigma_0$, then by bounded convergence theorem and $f(\sigma)>1$, we have
\[
1={\mathbb E}\left[\lim_{n\to\infty}e^{\sigma
M_{n\wedge\tau_1}} \left(\frac{1}{f(\sigma)}\right)^{n\wedge\tau_1}\right]
={\mathbb E}\left[1_{\{\tau_1<\infty\}}e^{\sigma} \left(\frac{1}{f(\sigma)}\right)^{\tau_1}\right].
\]
Let $\sigma\downarrow \sigma_0$, we get ${\mathbb P}(\tau_1<\infty)=e^{-\sigma_0}=\frac{p}{q}<1$.
\end{solution}

\noindent (iii) Compute ${\mathbb E}\alpha^{\tau_1}$ for $\alpha \in (0,1)$.

\begin{solution} Choose $\sigma > \sigma_0$, then $f(\sigma)>1$ and we have from (ii)
${\mathbb E}\left[1_{\{\tau_1<\infty\}}\left(\frac{1}{f(\sigma)}\right)^{\tau_1}\right]=e^{-\sigma}$. Set $\alpha=\frac{1}{f(\sigma)}$, then
\[
{\mathbb E}\left[\alpha^{\tau_1}1_{\{\tau_1<\infty\}}\right] = e^{-\sigma(\alpha)}
\]
where $\sigma(\alpha)$ satisfies $e^{\sigma(\alpha)}=\frac{1\pm\sqrt{1-4pq\alpha^2}}{2p\alpha}$. To make sure $\sigma(\alpha)>\sigma_0$,  we choose
$\sigma(\alpha)=\ln\left(\frac{1+\sqrt{1-4pq\alpha^2}}{2p\alpha}\right)$. As a consequence,
\[
{\mathbb E}\left[\alpha^{\tau_1}1_{\{\tau_1<\infty\}}\right]=\frac{1-\sqrt{1-4pq\alpha^2}}{2q\alpha}.
\]
\end{solution}

\noindent (iv) Compute ${\mathbb E}\left[1_{\{\tau_1<\infty\}}\tau_1\right]$. (Since ${\mathbb P}(\tau_1=\infty)>0$, we have ${\mathbb E}\tau_1=\infty$.)

\begin{solution} Using result in (iii) and bounded convergence theorem, we have
\begin{eqnarray*}
{\mathbb E}\left[\tau_11_{\{\tau_1<\infty\}}\right]
&=& {\mathbb E} \left[\lim_{\alpha\uparrow 1}\left(\tau_1\alpha^{\tau_1-1}\right)\right]
 =  \lim_{\alpha\uparrow 1} {\mathbb E} \left[\left(\tau_1\alpha^{\tau_1-1}\right)\right]
 =  \lim_{\alpha\uparrow 1}\frac{\partial}{\partial\alpha} {\mathbb E}[\alpha^{\tau_1}] \\
&=& \frac{1}{2q} \left[\frac{4pq}{\sqrt{1-4pq}}-(1-\sqrt{1-4pq})\right]=\frac{1}{2q}
\left[\frac{4pq}{2q-1}-1+2q-1\right] \\
&=& \frac{p}{q}\frac{1}{q-p}.
\end{eqnarray*}
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.4 (Distribution of $\tau_2$).} Consider the symmetric random walk, and let $\tau_2$ be the first time the random walk, starting from level 0, reaches the level 2. According to Theorem 5.2.3,
\[
{\mathbb E}\alpha^{\tau_2} = \left( \frac{1-\sqrt{1-\alpha^2}}{\alpha} \right)^2 \; \mbox{for all $\alpha\in (0,1)$.}
\]
Using the power series (5.2.21), we may write the right-hand side as
\begin{eqnarray*}
\left(\frac{1-\sqrt{1-\alpha^2}}{\alpha}\right)^2 &=& \frac{2}{\alpha} \cdot \frac{1-\sqrt{1-\alpha^2}}{\alpha} - 1 \\
&=& -1 + \sum_{j=1}^{\infty} \left(\frac{\alpha}{2}\right)^{2j-2} \frac{(2j-2)!}{j!(j-1)!} \\
&=& \sum_{j=2}^{\infty} \left(\frac{\alpha}{2}\right)^{2j-2} \frac{(2j-2)!}{j!(j-1)!} \\
&=& \sum_{k=1}^{\infty} \left(\frac{\alpha}{2}\right)^{2k} \frac{(2k)!}{(k+1)!k!}.
\end{eqnarray*}

\smallskip

\noindent (i) Use the power series above to determine ${\mathbb P}\left\{\tau_2=2k\right\}$, $k=1,2,\cdots$.

\begin{solution}
${\mathbb E}[\alpha^{\tau_2}]=\sum_{k=1}^{\infty} {\mathbb P}(\tau_2=2k)\alpha^{2k}=\sum_{k=1}^{\infty}\left(\frac{\alpha}{2}\right)^{2k} {\mathbb P}(\tau_2=2k)4^k$.
So ${\mathbb P}(\tau_2=2k)=\frac{(2k)!}{4^k(k+1)!k!}$.
\end{solution}

\noindent (ii) Use the reflection principle to determine ${\mathbb P}\{\tau_2=2k\}$, $k=1,2,\cdots$.

\begin{solution} ${\mathbb P}(\tau_2=2)=\frac{1}{4}$. For $k\ge 2$,
${\mathbb P}(\tau_2=2k)={\mathbb P}(\tau_2\le 2k)-{\mathbb P}(\tau_2\le 2k-2)$.
\begin{eqnarray*}
{\mathbb P}(\tau_2\le 2k)&=&{\mathbb P}(M_{2k}=2)+{\mathbb P}(M_{2k}\ge 4)+{\mathbb P}(\tau_2\le
2k,M_{2k}\le 0)\\
&=&{\mathbb P}(M_{2k}=2)+2{\mathbb P}(M_{2k}\ge 4)\\
&=&{\mathbb P}(M_{2k}=2)+{\mathbb P}(M_{2k}\ge 4)+{\mathbb P}(M_{2k}\le
-4)\\
&=&1-{\mathbb P}(M_{2k}=-2)-{\mathbb P}(M_{2k}=0).
\end{eqnarray*}
Similarly, ${\mathbb P}(\tau_2\le 2k-2)=1-{\mathbb P}(M_{2k-2}=-2)-{\mathbb P}(M_{2k-2}=0)$. So
\begin{eqnarray*}
{\mathbb P}(\tau_2=2k)&=&{\mathbb P}(M_{2k-2}=-2)+{\mathbb P}(M_{2k-2}=0)-{\mathbb P}(M_{2k}=-2)-{\mathbb P}(M_{2k}=0)\\
&=&\left(\frac{1}{2}\right)^{2k-2}\left[\frac{(2k-2)!}{k!(k-2)!} +
\frac{(2k-2)!}{(k-1)!(k-1)!}
\right]-\left(\frac{1}{2}\right)^{2k}\left[\frac{(2k)!}{(k+1)!(k-1)!}+\frac{(2k)!}{k!k!}\right]\\
&=&\frac{(2k)!}{4^k(k+1)!k!}\left[\frac{4}{2k(2k-1)}(k+1)k(k-1)+\frac{4}{2k(2k-1)}(k+1)k^2-k-(k+1)\right]\\
&=&\frac{(2k)!}{4^k(k+1)!k!}\left[\frac{2(k^2-1)}{2k-1}+\frac{2(k^2+k)}{2k-1}-\frac{4k^2-1}{2k-1}\right]\\
&=&\frac{(2k)!}{4^k(k+1)!k!}.
\end{eqnarray*}
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.5 (Joint distribution of random walk and maximum-to-date).} Let $M_n$ be a symmetric random walk, and define its {\it maximum-to-date} process
\[
M_n^* = \max_{1\le k \le n} M_k.\tag{5.7.2}
\]
Let $n$ and $m$ be even positive integers, and let $b$ be an even integer less than or equal to $m$. Assume $m\le n$ and $2m-b \le n$.

\smallskip

\noindent (i) Use an argument based on reflected paths to show that
\begin{eqnarray*}
{\mathbb P}\{M_n^* \ge m, M_n=b\} &=& {\mathbb P}\{M_n = 2m-b\} \\
&=& \frac{n!}{\left(\frac{n-b}{2}+m\right)!\left(\frac{n+b}{2}-m\right)!} \left(\frac{1}{2}\right)^n.
\end{eqnarray*}

\begin{proof} For every path that crosses level
$m$ by time $n$ and resides at $b$ at time $n$, there corresponds a
reflected path that resides at time $2m-b$. So
\[
{\mathbb P}(M_n^*\ge
m,\;M_n=b)={\mathbb P}(M_n=2m-b)=\left(\frac{1}{2}\right)^n\frac{n!}{\left(m+\frac{n-b}{2}\right)!\left(\frac{n+b}{2}-m\right)!}.
\]
\end{proof}

\noindent (ii) If the random walk is asymmetric with probability $p$ for an up step and probability $q=1-p$ for a down step, where $0<p<1$, what is ${\mathbb P}\{M_n^* \ge m, M_n=b\}$?

\begin{solution} The argument is similar to (i), but we need to determine the powers of $p$ and $q$ from the following equations
\[
\begin{cases}
x_p + x_q = n\\
x_p - x_q = 2m - b.
\end{cases}
\]
Solving it gives us
\[
\begin{cases}
x_p = m+\frac{n-b}{2}\\
x_q = \frac{n+b}{2}-m.
\end{cases}
\]
Therefore
\[
{\mathbb P}(M_n^*\ge
m,\;M_n=b)={\mathbb P}(M_n=2m-b)=\frac{n!}{\left(m+\frac{n-b}{2}\right)!\left(\frac{n+b}{2}-m\right)!}p^{m+\frac{n-b}{2}}q^{\frac{n+b}{2}-m}.
\]
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.6.} The value of the perpetual American put in Section 5.4 is the limit as $n\to\infty$ of the value of an American put with the same strike price 4 that expires at time $n$. When the initial stock price is $S_0=4$, the value of the perpetual American put is 1 (see (5.4.6) with $j=2$). Show that the value of an American put in the same model when the initial stock price is $S_0=4$ is 0.80 if the put expires at time 1, 0.928 if the put expires at time 3, and 0.96896 if the put expires at time 5.

\begin{proof} Instead of numerical examples, we give a rigorous proof.

On the infinite coin-toss space, we
define ${\cal M}_n=\{\mbox{stopping times that takes values
$0,1,\cdots,n,\infty$}\}$ and ${\cal M}_{\infty}=\{\mbox{stopping times
that takes values $0,1,2,\cdots$}\}$. Then the time-zero value $V^*$
of the perpetual American put as in Section 5.4 can be defined as
\[
V^* = \sup_{\tau\in {\cal M}_{\infty}}\widetilde
{\mathbb E} \left[1_{\{\tau<\infty\}}\frac{(K-S_{\tau})^+}{(1+r)^{\tau}}\right].
\]
For an
American put with the same strike price $K$ that expires at time
$n$, its time-zero value $V^{(n)}$ is
\[
V^{(n)} = \max_{\tau\in
{\cal M}_{n}}\widetilde
{\mathbb E} \left[1_{\{\tau<\infty\}}\frac{(K-S_{\tau})^+}{(1+r)^{\tau}}\right].
\]
Clearly
$(V^{(n)})_{n\ge 0}$ is nondecreasing and $V^{(n)}\le V^*$ for every
$n$. So $\lim_{n}V^{(n)}$ exists and $\lim_nV^{(n)}\le V^*$.

For any given $\tau\in {\cal M}_{\infty}$, we define $\tau^{(n)}=\left\{
\begin{matrix}
\infty,& \mbox{if $\tau=\infty$}\\
\tau\wedge n,& \mbox{if $\tau<\infty$}
\end{matrix}
\right.$,  then $\tau^{(n)}$ is also a stopping time, $\tau^{(n)}\in
{\cal M}_n$ and $\lim_{n\to\infty}\tau^{(n)}=\tau$. By bounded convergence
theorem,
\[
\widetilde
{\mathbb E}\left[1_{\{\tau<\infty\}}\frac{(K-S_{\tau})^+}{(1+r)^{\tau}}\right]=\lim_{n\to\infty}\widetilde
{\mathbb E}\left[1_{\{\tau^{(n)}<\infty\}}\frac{(K-S_{\tau^{(n)}})^+}{(1+r)^{\tau^{(n)}}}\right]\le
\lim_{n\to\infty}V^{(n)}.
\]Take $\sup$ at the left hand side of the inequality, we get $V^*\le
\lim_{n\to\infty}V^{(n)}$. Therefore $V^*=\lim_nV^{(n)}$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.7 (Hedging a short position in the perpetual American put).} Suppose you have sold the perpetual American put of Section 5.4 and are hedging the short position in this put. Suppose that at the current time the stock price is $s$ and the value of your hedging portfolio is $v(s)$. Your hedge is to first consume the amount\footnote{The text missed a factor of $\frac{1}{2}$ in fron of $v\left(\frac{s}{2}\right)$.}
\[
c(s) = v(s) - \frac{4}{5} \left[\frac{1}{2} v(2s) + \frac{1}{2} v\left(\frac{s}{2}\right)\right] \tag{5.7.3}
\]
and then take a position
\[
\delta(s) = \frac{v(2s)-v\left(\frac{s}{2}\right)}{2s - \frac{s}{2}} \tag{5.7.4}
\]
in the stock. (See Theorem 4.2.2 of Chapter 4. The processes $C_n$ and $\Delta_n$ in that theorem are obtained by replacing the dummy variable $s$ by the stock price $S_n$ in (5.7.3) and (5.7.4); i.e. $C_n = c(S_n)$ and $\Delta_n = \delta(S_n)$.) If you hedge this way, then regardless of whether the stock goes up or down on the next step, the value of your hedging portfolio should agree with the value of the perpetual American put.

\smallskip

\noindent (i) Compute $c(s)$ when $s=2^j$ for the three cases $j\le 0$, $j=1$, and $j\ge 2$.

\begin{solution} By (5.4.6),
\begin{eqnarray*}
v(2^j) =
\begin{cases}
4-2^j         & \mbox{if $j\le 1$} \\
\frac{4}{2^j} & \mbox{if $j\ge 1$.}
\end{cases}
\end{eqnarray*}
For $j\le 0$,
\[
c(2^j) = v(2^j) - \frac{4}{5}\left[\frac{1}{2}v(2^{j+1}) + \frac{1}{2}v(2^{j-1})\right] = (4-2^j) - \frac{2}{5}(4-2^{j+1}+4-2^{j-1}) = \frac{4}{5}.
\]
For $j=1$,
\[
c(2) = v(2) - \frac{4}{5} \left[\frac{1}{2}v(4) + \frac{1}{2}v(1)\right] = 2 - \frac{2}{5}(1+4-1) = \frac{2}{5}.
\]
For $j\ge 2$,
\[
c(2^j) = v(2^j) - \frac{2}{5}\left[v(2^{j+1})+v(2^{j-1})\right] = \frac{4}{2^j} - \frac{2}{5}\left(\frac{4}{2^{j+1}} + \frac{4}{2^{j-1}}\right) = 0.
\]
\end{solution}

\noindent (ii) Compute $\delta(s)$ when $s=2^j$ for the three cases $j\le 0$, $j=1$, and $j\ge 2$.

\begin{solution}
For $j\le 0$,
\[
\delta(2^j) = \frac{v(2^{j+1})-v(2^{j-1})}{2^{j+1}-2^{j-1}} = \frac{(4-2^{j+1})-(4-2^{j-1})}{2^{j+1}-2^{j-1}} = -1.
\]
For $j=1$,
\[
\delta(2) = \frac{v(4)-v(1)}{4-1}=\frac{1-(4-1)}{4-1} = -\frac{2}{3}.
\]
For $j\ge 2$,
\[
\delta(2^j) = \frac{v(2^{j+1})-v(2^{j-1})}{2^{j+1}-2^{j-1}} = \frac{\frac{4}{2^{j+1}}-\frac{4}{2^{j-1}}}{2^{j+1}-2^{j-1}} = -\frac{1}{4^{j-1}}.
\]
\end{solution}

\noindent (iii) Verify in each of the three cases $s=2^j$ for $j\le 0$, $j=1$, and $j\ge 2$ that the hedge works (i.e., regardless of whether the stock goes up or down, the value of your hedging portfolio at the next time is equal to the value of the perpetual American put at that time).

\begin{proof}
The computation is too tedious; skipped for this version.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.8 (Perpetual American call).} Like the perpetual American put of Section 5.4, the perpetual American call has no expiration. Consider a binomial model with up factor $u$, down factor $d$, and interest rate $r$ that satisfies the no-arbitrage condition $0<d<1+r<u$. The risk-neutral probabilities are
\[
\tilde p = \frac{1+r-d}{u-d}, \; \tilde q = \frac{u-1-r}{u-d}.
\]
The intrinsic value of the perpetual American call is $g(s) = s-K$, where $K>0$ is the strike price. The purpose of this exercise is to show that the value of the call is always the price of the underlying stock, and there is no optimal exercise time.

\smallskip

\noindent (i) Let $v(s)=s$. Show that $v(S_n)$ is always at least as large as the intrinsic value $g(S_n)$ of the call and $\left(\frac{1}{1+r}\right)^n v(S_n)$ is a supermartingale under the risk-neutral probabilities. In fact, $\left(\frac{1}{1+r}\right)^n v(S_n)$ is a martingale.\footnote{The textbook said ``supermartingale" by mistake.} These are the analogues of properties (i) and (ii) for the perpetual American put of Section 5.4.

\begin{proof}
$v(S_n)=S_n\ge S_n-K=g(S_n)$. Under
risk-neutral probabilities,
$\frac{1}{(1+r)^n}v(S_n)=\frac{S_n}{(1+r)^n}$ is a martingale by Theorem 2.4.4.
\end{proof}

\noindent (ii) to show that $v(s)=s$ is not too large to be the value of the perpetual American call, we must find a good policy for the purchaser of the call. Show that if the purchaser of the call exercises at time $n$, regardless of the stock price at that time, then the discounted risk-neutral expectation of her payoff is $S_0 - \frac{K}{(1+r)^n}$. Because this is true for every $n$, and
\[
\lim_{n\to \infty} \left[S_0 - \frac{K}{(1+r)^n}\right]=S_0,
\]
the value of the call at time zero must be at least $S_0$. (The same is true at all other times; the value of the call is at least as great as the current stock price.)

\begin{proof} If the purchaser chooses to exercises the call at
time $n$, then the discounted risk-neutral expectation of her payoff
is $\widetilde
{\mathbb E}\left[\frac{S_n-K}{(1+r)^n}\right]=S_0-\frac{K}{(1+r)^n}$. Since
$\lim_{n\to\infty}\left[S_0-\frac{K}{(1+r)^n}\right]=S_0$, the value
of the call at time zero is at least
$\sup_n\left[S_0-\frac{K}{(1+r)^n}\right]=S_0$.
\end{proof}

\noindent (iii) In place of (i) and (ii) above, we could verify that $v(s)=s$ is the value of the perpetual American call by checking that this function satisfies the equation (5.4.16) and boundary conditions (5.4.18). Do this verification.

\begin{proof}$\max\left\{g(s),\frac{\widetilde
pv(us)+\widetilde qv(ds)}{1+r}\right\}=\max\{s-K,\frac{\widetilde
pu+\widetilde qv}{1+r}s\}=\max\{s-K,s\}=s=v(s)$, so equation
(5.4.16) is satisfied. Clearly $v(s)=s$ also satisfies the boundary
condition (5.4.18).
\end{proof}

\noindent (iv) Show that there is no optimal time to exercise the perpetual American call.

\begin{proof} Suppose $\tau$ is an optimal exercise time, then according to result in (ii),
\[
\widetilde
{\mathbb E}\left[\frac{S_{\tau}-K}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right] \ge \sup_n\left[S_0-\frac{K}{(1+r)^n}\right] =
S_0,
\]
which implies ${\mathbb P}(\tau<\infty)\ne 0$ and hence $\widetilde
{\mathbb E}\left[\frac{K}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right]>0$. So
$\widetilde
{\mathbb E}\left[\frac{S_{\tau}-K}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right]<\widetilde
{\mathbb E}\left[\frac{S_{\tau}}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right]$.
Since $\left(\frac{S_n}{(1+r)^n}\right)_{n\ge 0}$ is a martingale
under the risk-neutral measure, by Fatou's lemma,
\[
\widetilde
{\mathbb E}\left[\frac{S_{\tau}}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right]
\le
\liminf_{n\to\infty}\widetilde {\mathbb E}\left[\frac{S_{\tau\wedge
n}}{(1+r)^{\tau\wedge
n}}1_{\{\tau<\infty\}}\right]
\le \liminf_{n\to\infty}\widetilde
{\mathbb E}\left[\frac{S_{\tau\wedge n}}{(1+r)^{\tau\wedge
n}}\right]=\liminf_{n\to\infty}\widetilde {\mathbb E}[S_0]=S_0.
\]
Combined, we
have
\[
S_0 \le \widetilde
{\mathbb E}\left[\frac{S_{\tau}-K}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right]<\widetilde
{\mathbb E}\left[\frac{S_{\tau}}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right] \le S_0.
\]
Contradiction. So there is no optimal time to exercise the perpetual
American call. Simultaneously, we have shown $\widetilde
{\mathbb E}\left[\frac{S_{\tau}-K}{(1+r)^{\tau}}1_{\{\tau<\infty\}}\right]<S_0$
for any stopping time $\tau$. Combined with (ii), we conclude
$S_0$ is the least upper bound for all the prices acceptable to the
buyer.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 5.9.} (Provided by Irene Villegas.) Here is a method for solving equation (5.4.13) for the value of the perpetual American put in Section 5.4.

\smallskip

\noindent (i) We first determine $v(s)$ for large values of $s$. When $s$ is large, it is not optimal to exercise the put, so the maximum in (5.4.13) will be given by the second term
\[
\frac{4}{5} \left[ \frac{1}{2}v(2s) + \frac{1}{2}v\left(\frac{s}{2}\right) \right] = \frac{2}{5}v(2s)+\frac{2}{5}v\left(\frac{s}{2}\right).
\]
We thus seek solutions to the equation
\[
v(s) = \frac{2}{5} v(2s) + \frac{2}{5}v\left(\frac{s}{2}\right). \tag{5.7.5}
\]
All such solutions are of the form $s^p$ for some constant $p$ or linear combination of functions of this form. Substitute $s^p$ into (5.7.5), obtain a quadratic equation for $2^p$, and solve to obtain $2^p = 2$ or $2^p = \frac{1}{2}$. This leads to the values $p=1$ and $p=-1$, i.e., $v_1(s)=s$ and $v_2(s)=\frac{1}{2}$ are soltuions to (5.7.5).

\begin{proof}
Suppose $v(s)=s^p$, then we have
$s^p=\frac{2}{5}2^ps^p+\frac{2}{5}\frac{s^p}{2^p}$. So
$1=\frac{2^{p+1}}{5}+\frac{2^{1-p}}{5}$. Solve it for $p$, we get
$p=1$ or $p=-1$.\end{proof}

\noindent (ii) The general solution to (5.7.5) is a linear combination of $v_1(s)$ and $v_2(s)$, i.e.,
\[
v(s) = As + \frac{B}{s}. \tag{5.7.6}
\]
For large values of $s$, the value of the perpetual American put must be given by (5.7.6). It remains to evaluate $A$ and $B$. Using the second boundary condition in (5.4.15), show that $A$ must be zero.

\begin{proof}Since
$\lim_{s\to\infty}v(s)=\lim_{s\to\infty}(As+\frac{B}{s})=0$, we must
have $A=0$.\end{proof}

\noindent (iii) We have thus established that for large values of $s$, $v(s)=\frac{B}{s}$ for some constant $B$ still to be determined. For small values of $s$, the value of the put is its intrinsic value $4-s$. We must choose $B$ so these two functions coincide at some point, i.e., we must find a value for $B$ so that, for some $s>0$,
\[
f_B(s) = \frac{B}{s} - (4-s)
\]
equals zero. Show that, when $B>4$, this function does not take the value $0$ for any $s>0$, but, when $B\le 4$, the equation $f_B(s)=0$ has a solution.

\begin{proof}$f_B(s)=0$ if and only if $B+s^2-4s=0$. The
discriminant $\Delta=(-4)^2-4B=4(4-B)$. So for $B\le 4$, the
equation has roots and for $B>4$, this equation does not have roots.
\end{proof}

\noindent (iv) Let $B$ be less than or equal to $4$, and let $s_B$ be a solution of the equation $f_B(s)=0$. Suppose $s_B$ is a stock price that can be attained in the model (i.e., $s_B=2^j$ for some integer $j$). Suppose further that the owner of the perpetual American put exercises the first time the stock price is $s_B$ or smaller. Then the discounted risk-neutral expected payoff of the put is $v_B(S_0)$, where $v_B(s)$ is given by the formula
\[
v_B(s) =
\begin{cases}
4 - s, & \mbox{if $s\le s_B$,} \\
\frac{B}{s}, & \mbox{if $s\ge s_B$.}
\end{cases}
\tag{5.7.7}
\]
Which values of $B$ and $s_B$ give the owner the largest option value?
\begin{proof}Suppose $B\le 4$, then the equation $s^2-4s+B=0$
has solution $2\pm\sqrt{4-B}$. By drawing graphs of $4-s$ and
$\frac{B}{s}$, we can see the largest $v_B(s)$ is obtained by making $4-s$ and $\frac{B}{s}$ tangent to each other, which implies
\[
\begin{cases}
\Delta = 4(4-B) = 0 \\
4-s_B = \frac{B}{s_B}.
\end{cases}
\]
Solving it gives us $B=4$ and $s_B=2$.
\end{proof}

\noindent (v) For $s<s_B$, the derivative of $v_B(s)$ is $v_B'(s) = -1$. For $s>s_B$, this derivative is $v_B'(s)=-\frac{B}{s^2}$. Show that the best value of $B$ for the option owner makes the derivative of $v_B(s)$ continuous at $s=S_B$ (i.e., the two formulas for $v_B'(s)$ give the same answer at $s=s_B$).

\begin{proof}To have continuous derivative, we must have
$-1=-\frac{B}{s_B^2}$. Plug $B=s_B^2$ back into $4-s_B = \frac{B}{s_B}$, we
get $s_B=2$. This gives $B=4$.
\end{proof}


\section{Interest-Rate-Dependent Assets}


$\bigstar$ {\bf Comments}:

\medskip

1) In previous chapters, we started with a real probability measure and derived the risk-neutral measure based on the no-arbitrage argument. The concrete form of the risk-neutral measure is known in that setting. In the current chapter, models are built under the risk-neutral measure, whose existence is assumed {\it a priori} and whose concrete form (in terms of the real probability measure) is unknown. The justification for this is the following result by Dalang et al. \cite{DMW89}:
\begin{theorem}[Dalang-Morton-Willinger, {\it Fundamental Theorem of Asset Pricing}]
Let $(\Omega, {\cal F}, \{{\cal F}_t\}_{t=0,\cdots,T}, {\mathbb P})$ be a (general) filtered probability space and let $S = \{S_t\}_{t\in \{0,\cdots,T\}}$ be an adapted, ${\mathbb R}^d$-valued stochastic process describing the discounted prices of $d \in {\mathbb N}$ financial assets. Then the following properties are equivalent:

\noindent (a) The financial market model is free of arbitrage.

\noindent (b) There exists a probability measure ${\mathbb P}^*$ on $(\Omega, {\cal F}, {\mathbb P})$ such that

$\bullet$ ${\mathbb P}^*$ is equivalent to ${\mathbb P}$ and the Radon-Nikod\'{y}m density $\varrho:=d{\mathbb P}^*/d{\mathbb P}$ is bounded, i.e. in $L^{\infty}(\Omega, {\cal F}_T, {\mathbb P})$,

$\bullet$ Integrability: $S_t \in L^1(\Omega, {\cal F}_t, {\mathbb P}^*)$ for all $t \in \{0, \cdots, T\}$,

$\bullet$ Martingale property w.r.t. ${\mathbb P}$:
\[
{\mathbb E}_{{\mathbb P}^*}[S_t | {\cal F}_{t-1}] \overset{a.s.}{=} S_{t-1} \; \mbox{for all $t \in \{0, \cdots, T\}$.}
\]
\end{theorem}

According to the {\it Fundamental Theorem of Asset Pricing} (FTAP), the no-arbitrage property is associated with the existence of a probability measure called {\it equivalent martingale measure} (EMM) and a positive process called  {\it num\'{e}raire}, such that the price processes of tradable assets discounted by the num\'{e}raire are martingales under the equivalent martingale measure. In the current chapter, the num\'{e}raire is the value process of the money market account
\[
M_n = (1+R_0)\cdots(1+R_{n-1}), \; n=1, 2, \cdots, N; M_0 = 1.
\]
And the risk-neutral measure is the EMM associated with this num\'{e}raire. This justifies Definition 6.2.4 (zero-coupon bond prices) since
\[
D_n B_{n,m} = \widetilde {\mathbb E}_n \left[D_m B_{m,m}\right] \Rightarrow B_{n,m} = \widetilde {\mathbb E}_n \left[\frac{D_m \cdot 1}{D_n }\right].
\]

\medskip

2) The wealth equation (6.2.6) essentially assumes all coupons and other payouts of cash are reinvested in the portfolio.

\medskip

3) To see the intuition of the forward interest rate at time $n$ for investing at time $m$ (Definition 6.3.4), note at time $n$, we can invest \$1 to buy $\frac{1}{B_{n,m}}$ shares of zero-coupon bond maturing at time $m$; at time $m$, we reinvest the payoff $\frac{1}{B_{n,m}}$ at rate $F_{n,m}$ to get paid at time $(m+1)$. To eliminate arbitrage, this investing strategy should have the same return as investing \$1 at time $n$ in the zero-coupon bond maturing at time $(m+1)$:
\[
\frac{1}{B_{n,m}} (1+F_{n,m}) = \frac{1}{B_{n,m+1}}.
\]
This gives formula (6.3.3)
\[
F_{n,m} = \frac{B_{n,m}}{B_{n,m+1}} - 1 = \frac{B_{n,m}-B_{n,m+1}}{B_{n,m+1}}.
\]

\medskip

4) Whenever there is a hedge, static or not, the discounted value of the hedging portfolio is a martingale under the risk-neutral measure, and the risk-neutral pricing formula applies. This is regardless of whether or not the market is complete.

i) {\it Forward contract}. a) Hedging of a short position: at time $n$, short $\frac{S_n}{B_{n,m}}$ zero-coupon bond maturing at time $m$ and long 1 share of asset at price of $S_n$, hold the portfolio until maturity time $m$. b) Risk-neutral pricing:
\[
\widetilde {\mathbb E}_n \left[D_m (K-S_m) \right] = K D_n B_{n,m} - D_nS_n = D_n V_n,
\]
where $V_n$ is the contract value at time $n$. By setting $V_n=0$, we obtain $m$-forward price of the asset at time $n$ (Theorem 6.3.2).

ii) {\it Forward rate agreement} (FRA).\footnote{A FRA is a contract involving three time instants: the current time $t$, the expiry time $T>t$, and the maturity time $S>T$. The contract gives its holder an interest-rate payment for the period between $T$ and $S$. See Brigo and Mercurio \cite[page~11]{BM07} for a general definition.} a) Hedging of a short position: the portfolio whose value is $R_m$ at time $(m+1)$ can be replicated by a cash inflow of \$1 at time $m$ and a cash outflow of \$1 at time $(m+1)$, since the inflow of \$1 at time $m$ can be invested in the money market account to return $(1+R_m)$ at time $(m+1)$. The cash inflow of \$1 at time $m$ can be obtained by longing at time $n$ a zero-coupon bond maturing at time $m$, while the cash outflow of \$1 at time $(m+1)$ can be obtained by shorting at time $n$ a zero-coupon bond maturing at time $(m+1)$. So the value of the portfolio at time $n$ is
\[
V_n = B_{n,m} - B_{n,m+1}.
\]
b) Risk-neutral pricing (Exercise 6.3):
\[
V_n = \widetilde {\mathbb E}_n \left[\frac{D_{m+1}R_m}{D_n}\right] = \widetilde {\mathbb E}_n \left[\frac{D_m - D_{m+1}}{D_n}\right] =  B_{n,m} - B_{n,m+1}.
\]
This is Theorem 6.3.5.

iii) {\it Interest rate swap}. A long swap position (``receive fixed") is a long position in a coupon bond plus a short position in a series of FRA's. Therefore, an interest rate swap can be hedged and risk-neutral pricing formula applies. This is Theorem 6.3.7. More specifically,
\[
Swap_m = K \sum_{n=1}^m + B_{0,m} - 1
\]
suggests a long position in a bond with coupon $K$ and maturity $m$ and a short position of \$1 in the money market account. A short position of \$1 in the money market account at the beginning of each period will result in $-(1+R_n)$ at the end of the period.

iv) {\it Interest rate caps and floors}. Suppose the replication uses the money market account and zero-coupon bonds with various maturities. The wealth equation satisfies
\[
X_{n+1} = (1+R_n)(X_n - \Delta_n B_{n,m}) + \Delta_n B_{n+1,m}.
\]
To successfully replicate $V_{n+1}$, it's necessary and sufficient that $B_{n+1,m}(H)\ne B_{n+1,m}(T)$. See Exercise 6.4 for illustration.

\medskip

5) According to the {\it Fundamental Theorem of Asset Pricing} (FTAP), the no-arbitrage property is associated with the existence of a probability measure called {\it equivalent martingale measure} (EMM) and a positive process called  {\it num\'{e}raire}, such that the price processes of tradable assets discounted by the num\'{e}raire are martingales under the given probability measure (hence the name {\it martingale measure}).

The {\it $m$-forward measure} is the EMM associated with the zero-coupon bond $(B_{n,m})_{n=0}^m$ as the num\'{e}raire. For a European contingent claim with maturity $m$, we have the risk-neutral pricing formula (note $B_{m,m}\equiv 1$)
\[
V_n = B_{n,m} \widetilde {\mathbb E}^m_n[V_m] = \frac{1}{D_n} \widetilde {\mathbb E}_n[D_m V_m].
\]
If we denote the Radon-Nikod\'{y}m derivative process by $(Z_{n,m})_{n=0}^m$, that is, 
\[
Z_{m, m} := \frac{d\widetilde {\mathbb P}^m}{d\widetilde {\mathbb P}}, \; Z_{n, m} := \widetilde {\mathbb E}_n[Z_{m,m}], 
\]
by setting $n=0$ in the risk-neutral pricing formula, we have $Z_{m,m} = \frac{D_m}{B_{0,m}}$ and hence
\[
Z_{n,m} = \frac{\widetilde {\mathbb E}_n[D_m]}{B_{0,m}} = \frac{D_n B_{n,m}}{B_{0,m}}, \; n=0, \cdots, m.
\]
This gives an alternative presentation of the $m$-forward measure $\widetilde {\mathbb P}^m$. A notable feature of working under this measure is that any asset price process discounted by the zero-coupon bond $(B_{n,m})_{n=0}^m$ is exactly its $m$-forward price process (Theorem 6.3.2).

\bigskip

\noindent $\blacktriangleright$ {\bf Exercise 6.1.} Prove Theorem 2.3.2 when conditional expectation is defined by Definition 6.2.2.

\begin{proof}
The proof is tedious and is not a lot simpler than the one using the most general definition of conditional expectation (see, for example, Shiryaev \cite{Shiryaev95}); skipped for this version.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 6.2.} Verify that the discounted value of the static hedging portfolio constructed in the proof of Theorem 6.3.2 is a martingale under $\widetilde {\mathbb P}$.

\begin{proof} The static hedging portfolio consists of shorting one forward contract, shorting $\frac{S_n}{B_{n,m}}$ zero-coupon bonds maturing at time $m$, and longing one share of the asset. So for any $k = n, \cdots, m$, the value of the portfolio at time $k$ is
\[
X_k=S_k-\widetilde {\mathbb E}_k[D_m(S_m-K)]D_k^{-1}-\frac{S_n}{B_{n,m}}B_{k,m}.
\]
Then
\begin{eqnarray*}
\widetilde {\mathbb E}_{k-1}[D_kX_k]&=&\widetilde {\mathbb E}_{k-1}\left[D_kS_k-\widetilde {\mathbb E}_k[D_m(S_m-K)]-\frac{S_n}{B_{n,m}}B_{k,m}D_k\right]\\
&=&D_{k-1}S_{k-1}-\widetilde {\mathbb E}_{k-1}[D_m(S_m-K)]-\frac{S_n}{B_{n,m}}\widetilde {\mathbb E}_{k-1}\left[\widetilde {\mathbb E}_k[D_m\cdot 1]\right]\\
&=&D_{k-1}\left\{S_{k-1}-\widetilde {\mathbb E}_{k-1}[D_m(S_m-K)]D^{-1}_{k-1}-\frac{S_n}{B_{n,m}}B_{k-1,m}\right\}\\
&=&D_{k-1}X_{k-1}.
\end{eqnarray*}
This shows the discounted value of the static hedging portfolio constructed in the proof of Theorem 6.3.2 is a martingale under $\widetilde {\mathbb P}$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 6.3.} Let $0 \le n \le m \le N-1$ be given. According to the risk-neutral pricing formula, the contract that pays $R_m$ at time $m+1$ has time-$n$ price $\frac{1}{D_n}\widetilde {\mathbb E}_n[D_{m+1}R_m]$. Use the properties of conditional expectations to show that this gives the same result as Theorem 6.3.5, i.e.,
\[
\frac{1}{D_n} \widetilde {\mathbb E}_n[D_{m+1}R_m] = B_{n,m}-B_{n,m+1}.
\]

\begin{proof}
\[
\frac{1}{D_n}\widetilde {\mathbb E}_n[D_{m+1}R_m]=\frac{1}{D_n}\widetilde
{\mathbb E}_n[D_m(1+R_m)^{-1}R_m]= \frac{1}{D_n} \widetilde
{\mathbb E}_n\left[D_m-D_{m+1}\right]=B_{n,m}-B_{n,m+1}.
\]
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 6.4 (Short position hedge for caplets).} Using the data in Example 6.3.9, this exercise constructs a hedge for a short position in the caplet paying $\left(R_2-\frac{1}{3}\right)^+$ at time three. We observe from the second table in Example 6.3.9 that the payoff at time three of this caplet is
\[
V_3(HH) = \frac{2}{3}, \; V_3(HT) = V_3(TH) = V_3(TT) = 0.
\]
Since this payoff depends on only the first two coin tosses, the price of the caplet at time two can be determined by discounting:
\[
V_2(HH) = \frac{1}{1+R_2(HH)}V_3(HH)=\frac{1}{3}, \; V_2(HT) = V_2(TH) = V_2(TT)=0.
\]
Indeed, if one is hedging a short position in the caplet and has a portfolio valued at $\frac{1}{3}$ at time two in the event $\omega_1=H$, $\omega_2=H$, then one can simply invest this $\frac{1}{3}$ in the money market in order to have the $\frac{2}{3}$ required to pay off the caplet at time three.

In Example 6.3.9, the time-zero price of the caplet is determined to be $\frac{2}{21}$ (see (6.3.10)).

\smallskip

\noindent (i) Determine $V_1(H)$ and $V_1(T)$, the price at time one of the caplet in the events $\omega_1=H$ and $\omega_1=T$, respectively.

\begin{solution}
$D_1V_1=\widetilde {\mathbb E}_1[D_2V_2]=D_2\widetilde {\mathbb E}_1[V_2]$. So
$V_1=\frac{D_2}{D_1}\widetilde {\mathbb E}_1[V_2]=\frac{1}{1+R_1}\widetilde {\mathbb E}_1[V_2]$. In
particular,
\begin{eqnarray*}
& & V_1(H)=\frac{V_2(HH)P(\omega_2=H|\omega_1=H)+
V_2(HT)P(\omega_2=T|\omega_1=H)}{1+R_1(H)}
=\frac{6}{7}\left(\frac{1}{3}\cdot\frac{2}{3}+0\cdot\frac{1}{3}\right)=\frac{4}{21}, \\
& & V_1(T)=\frac{V_2(TH)P(\omega_2=H|\omega_1=T)+
V_2(TT)P(\omega_2=T|\omega_1=T)}{1+R_1(T)}=0.
\end{eqnarray*}
\end{solution}

\noindent (ii) Show how to begin with $\frac{2}{21}$ at time zero and invest in the money market and the maturity two bond in order to have a portfolio value $X_1$ at time one that agrees with $V_1$, regardless of the outcome of the first coin toss. Why do we invest in the maturity two bond rather than the maturity three bond to do this?

\begin{proof}Let $X_0=\frac{2}{21}$. Suppose we buy $\Delta_0$
shares of the maturity two bond, then at time one, the value of our
portfolio is $X_1=(1+R_0)(X_0-\Delta_0 B_{0,2})+\Delta_0 B_{1,2}$. To
replicate the value $V_1$, we must have
\begin{eqnarray*}
\left\{
\begin{matrix}
V_1(H)=(1+R_0)(X_0-\Delta_0B_{0,2})+\Delta_0B_{1,2}(H)\\
V_1(T)=(1+R_0)(X_0-\Delta_0B_{0,2})+\Delta_0B_{1,2}(T).
\end{matrix}
\right.
\end{eqnarray*}
So
\[
\Delta_0=\frac{V_1(H)-V_1(T)}{B_{1,2}(H)-B_{1,2}(T)}=\frac{\frac{4}{21}-0}{\frac{6}{7}-\frac{5}{7}}=\frac{4}{3}.
\]
The hedging strategy is therefore to borrow
$\frac{4}{3}B_{0,2}-\frac{2}{21}=\frac{20}{21}$ and buy
$\frac{4}{3}$ share of the maturity two bond. The reason why we do
not invest in the maturity three bond is that
$B_{1,3}(H)=B_{1,3}(T)(=\frac{4}{7})$ and the portfolio will
therefore have the same value at time one regardless the outcome of
first coin toss. This makes impossible the replication of $V_1$,
since $V_1(H)\ne V_1(T)$.
\end{proof}

\noindent (iii) Show how to take the portfolio value $X_1$ at time one and invest in the money market and the maturity three bond in order to have a portfolio value $X_2$ at time two that agrees with $V_2$, regardless of the outcome of the first two coin tosses. Why do we invest in the maturity three bond rather than the maturity two bond to do this?

\begin{proof}Suppose we buy $\Delta_1$ share of the maturity
three bond at time one, then to replicate $V_2$ at time two, we must
have $V_2=(1+R_1)(X_1-\Delta_1B_{1,3})+\Delta_1B_{2,3}$. So
\[
\Delta_1(H)=\frac{V_2(HH)-V_2(HT)}{B_{2,3}(HH)-B_{2,3}(HT)}=-\frac{2}{3}, \; \Delta_1(T)=\frac{V_2(TH)-V_2(TT)}{B_{2,3}(TH)-B_{2,3}(TT)}=0.
\]
The hedging strategy is as follows. If the outcome of first coin
toss is $T$, then we keep everything in the money market account. If the outcome of first coin toss
is $H$, we short $\frac{2}{3}$ shares of the maturity three bond
and invest the income into the money market account. We do not
invest in the maturity two bond, because at time two, the value of
the bond is its face value and our portfolio will therefore have the
same value regardless outcomes of coin tosses. This makes impossible
the replication of $V_2$.
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 6.5.} Let $m$ be given with $0\le m \le N-1$, and consider the forward interest rate
\[
F_{n,m} = \frac{B_{n,m}-B_{n,m+1}}{B_{n,m+1}}, \; n = 0, 1, \cdots, m.
\]

\smallskip

\noindent (i) Use (6.4.8) and (6.2.5) to show that $F_{n,m}$, $n=0, 1, \cdots, m$, is a martingale under the $(m+1)$-forward measure $\widetilde {\mathbb P}^{m+1}$.

\begin{proof} Before we start the computation, note $F_{n,m} = \frac{B_{n,m} - B_{n,m+1}}{B_{n,m+1}}$ is a tradable asset discounted by $(B_{n,m+1})_n$, so it must be a martingale under the equivalent martingale measure $\widetilde {\mathbb P}^{m+1}$ associated with the num\'{e}raire $(B_{n,m+1})_n$. To verify this, suppose $1\le n\le m$, then
\begin{eqnarray*}
\widetilde {\mathbb E}_{n-1}^{m+1}[F_{n,m}] 
&=&\widetilde
{\mathbb E}_{n-1}[B^{-1}_{n,m+1}(B_{n,m}-B_{n,m+1})Z_{n,m+1}Z^{-1}_{n-1,m+1}]\\
&=&\widetilde
{\mathbb E}_{n-1}\left[\left(\frac{B_{n,m}}{B_{n,m+1}}-1\right)\frac{B_{n,m+1}D_n}{B_{n-1,m+1}D_{n-1}}\right]\\
&=&\frac{D_n}{B_{n-1,m+1}D_{n-1}}\widetilde
{\mathbb E}_{n-1}[B_{n,m}-B_{n,m+1}]\\
&=&\frac{D_n}{B_{n-1,m+1}D_{n-1}}\widetilde
{\mathbb E}_{n-1}[D_n^{-1}\widetilde {\mathbb E}_n[D_m]-D_n^{-1}\widetilde
{\mathbb E}_n[D_{m+1}]]\\
&=&\frac{\widetilde {\mathbb E}_{n-1}[D_m-D_{m+1}]}{B_{n-1,m+1}D_{n-1}}\\
&=&\frac{B_{n-1,m}-B_{n-1,m+1}}{B_{n-1,m+1}}\\
&=&F_{n-1,m}.
\end{eqnarray*}
\end{proof}

\noindent (ii) Compute $F_{0,2}$, $F_{1,2}(H)$, and $F_{1,2}(T)$ in Example 6.4.4 and verify the martingale property
\[
\widetilde {\mathbb E}^3[F_{1,2}] = F_{0,2}.
\]
\begin{solution}
\begin{eqnarray*}
& & F_{0,2} = \frac{B_{0,2}}{B_{0,3}} - 1 = \frac{0.9071}{0.8639} - 1 = 0.05 \\
& & F_{1,2}(H) = \frac{B_{1,2}(H)}{B_{1,3}(H)} - 1 = \frac{0.9479}{0.8985} - 1 = 0.055 \\
& & F_{1,2}(T) = \frac{B_{1,2}(T)}{B_{1,3}(T)} - 1 = \frac{0.9569}{0.9158} - 1 = 0.045 \\
\end{eqnarray*}
Therefore
\[
\widetilde {\mathbb E}^3[F_{1,2}] 
= \widetilde{\mathbb P}^3(H) F_{1,2}(H) + \widetilde{\mathbb P}^3(T) F_{1,2}(T)
= 0.4952 \cdot  0.055 + 0.5048 \cdot 0.045 = 0.05 = F_{0,2}.
\]
\end{solution}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 6.6.} Let $S_m$ be the price at time $m$ of an asset in a binomial interest rate model. For $n=0,1,\cdots,m$, the forward price is $\mbox{For}_{n,m} = \frac{S_n}{B_{n,m}}$ and the futures price is $\mbox{Fut}_{n,m} = \widetilde{\mathbb E}_n[S_m]$.\footnote{The textbook said ``$S_n$" by mistake.}

\smallskip

\noindent (i) Suppose that at each time $n$ an agent takes a long forward position and sells this contract at time $n+1$. Show that this generates cash flow $S_{n+1}-\frac{S_nB_{n+1,m}}{B_{n,m}}$ at time $n+1$.

\begin{proof}
At time $n+1$, the
forward contract has a value of
\[
\frac{1}{ D_{n+1}} \widetilde
{\mathbb E}_{n+1} \left[D_m(S_m-\mbox{Fut}_{n,m})\right] = S_{n+1}-\mbox{Fut}_{n,m} B_{n+1,m}=S_{n+1}-\frac{S_nB_{n+1,m}}{B_{n,m}}.
\]
So if the agent takes a long forward position at time $n$ for no cost and sells this contract at time $n+1$, he will receive a
cash flow of $S_{n+1}-\frac{S_nB_{n+1,m}}{B_{n,m}}$.
\end{proof}

\noindent (ii) Show that if the interest rate is a constant $r$ and at each time $n$ an agent takes a long position of $(1+r)^{m-n-1}$ forward contracts, selling these contracts at time $n+1$, then the resulting cash flow is the same as the difference in the futures price $\mbox{Fut}_{n+1,m} - \mbox{Fut}_{n,m}$.

\begin{proof} By (i), the cash flow generated at time $n+1$ is
\begin{eqnarray*}
&
&(1+r)^{m-n-1}\left(S_{n+1}-\frac{S_nB_{n+1,m}}{B_{n,m}}\right)\\
&=&(1+r)^{m-n-1}\left(S_{n+1}-\frac{\frac{S_n}{(1+r)^{m-n-1}}}{\frac{1}{(1+r)^{m-n}}}\right)\\
&=&(1+r)^{m} \frac{S_{n+1}}{(1+r)^{n+1}}-(1+r)^{m}\frac{S_n}{(1+r)^n}\\
&=&(1+r)^m\widetilde {\mathbb E}_{n+1} \left[\frac{S_m}{(1+r)^m}\right]-(1+r)^m\widetilde
{\mathbb E}_n \left[\frac{S_m}{(1+r)^m}\right]\\
&=& \mbox{Fut}_{n+1,m}-\mbox{Fut}_{n,m}.
\end{eqnarray*}
\end{proof}

\medskip

\noindent $\blacktriangleright$ {\bf Exercise 6.7.} Consider a binomial interest rate model in which the interest rate at time $n$ depends on only the number of heads in the first $n$ coin tosses. In other words, for each $n$ there is a function $r_n(k)$ such that
\[
R_n(\omega_1,\cdots,\omega_n)=r_n(\#H(\omega_1,\cdots,\omega_n)).
\]
Assume the risk-neutral probabilities are $\tilde p = \tilde q = \frac{1}{2}$. The Ho-Lee model (Example 6.4.4) and the Black-Derman-Toy model (Example 6.5.5) satisfy these conditions.

Consider a derivative security that pays 1 at time $n$ if and only if there are $k$ heads in the first $n$ tosses; i.e., the payoff is $V_n(k) = 1_{\{\#H(\omega_1,\cdots,\omega_n)=k\}}$. Define $\psi_0(0)=1$ and , for $n=1,2,\cdots$, define
\[
\psi_n(k) = \widetilde{\mathbb E}\left[D_nV_n(k)\right], \; k = 0, 1, \cdots, n,
\]
to be the price of this security at time zero. Show that the functions $\psi_n(k)$ can be computed by the recursion
\begin{eqnarray*}
& & \psi_{n+1}(0) = \frac{\psi_n(0)}{2(1+r_n(0))} \\
& & \psi_{n+1}(k) = \frac{\psi_n(k-1)}{2(1+r_n(k-1))} + \frac{\psi_n(k)}{2(1+r_n(k))}, \; k=1, \cdots, n,\\
& & \psi_{n+1}(n+1) = \frac{\psi_n(n)}{2(1+r_n(n))}.
\end{eqnarray*}

 \begin{proof}
\begin{eqnarray*}\psi_{n+1}(0)&=&\widetilde
{\mathbb E} \left[D_{n+1}V_{n+1}(0)\right]\\
&=&\widetilde {\mathbb E} \left[\frac{D_n}{1+r_n(0)}1_{\{\#H(\omega_1\cdots
\omega_{n+1})=0\}}\right]\\
&=&\widetilde {\mathbb E} \left[\frac{D_n}{1+r_n(0)}1_{\{\#H(\omega_1\cdots
\omega_{n})=0\}}1_{\{\omega_{n+1}=T\}}\right]\\
&=&\frac{1}{2(1+r_n(0))}\widetilde
{\mathbb E} \left[D_n1_{\{\#H(\omega_1\cdots
\omega_{n})=0\}}\right]\\
&=&\frac{\psi_n(0)}{2(1+r_n(0))}.
\end{eqnarray*}
For $k=1,2,\cdots,n$, 
\begin{eqnarray*} 
\psi_{n+1}(k)&=&\widetilde
{\mathbb E}\left[\frac{D_n}{1+r_n(\#H(\omega_1\cdots\omega_n))}1_{\{\#H(\omega_1\cdots
\omega_{n+1})=k\}}\right]\\
&=&\widetilde {\mathbb E}\left[\frac{D_n}{1+r_n(k)}1_{\{\#H(\omega_1\cdots
\omega_{n})=k\}}1_{\{\omega_{n+1}=T\}}\right]+\widetilde
{\mathbb E}\left[\frac{D_n}{1+r_n(k-1)}1_{\{\#H(\omega_1\cdots
\omega_{n})=k-1\}}1_{\{\omega_{n+1}=H\}}\right]\\
&=&\frac{1}{2}\frac{\widetilde
{\mathbb E}[D_nV_n(k)]}{1+r_n(k)}+\frac{1}{2}\frac{\widetilde
{\mathbb E}[D_nV_n(k-1)]}{1+r_n(k-1)}\\
&=&\frac{\psi_n(k)}{2(1+r_n(k))}+\frac{\psi_n(k-1)}{2(1+r_n(k-1))}.
\end{eqnarray*}
Finally,
\[
\psi_{n+1}(n+1)=\widetilde {\mathbb E}[D_{n+1}V_{n+1}(n+1)]=\widetilde
{\mathbb E}\left[\frac{D_n}{1+r_n(n)}1_{\{\#H(\omega_1\cdots
\omega_{n})=n\}}1_{\{\omega_{n+1}=H\}}\right]=\frac{\psi_n(n)}{2(1+r_n(n))}.
\]
\end{proof}

\begin{remark} In the above proof, we have used the independence of
$\omega_{n+1}$ and $(\omega_1,\cdots,\omega_n)$ under $\widetilde{\mathbb P}$. This is guaranteed
by the assumption that $\widetilde{\mathbb P}(\omega_{n+1}=H|\omega_1,\cdots,\omega_n)=\tilde p$ and $\widetilde{\mathbb P}(\omega_{n+1}=T|\omega_1,\cdots,\omega_n)=\tilde q$, which are deterministic. In case the
binomial model has stochastic up- and down-factor $u_n$ and $d_n$,
we have $\widetilde
{\mathbb P}(\omega_{n+1}=H|\omega_1,\cdots,\omega_n)=\tilde p_n$ and $\widetilde
{\mathbb P}(\omega_{n+1}=T|\omega_1,\cdots,\omega_n)=\tilde q_n$, where
$\tilde p_n=\frac{1+r_n-d_n}{u_n-d_n}$ and $\tilde q_n=\frac{u_n-1-r_n}{u_n-d_n}$. Then for any
$X\in {\cal F}_n=\sigma(\omega_1,\cdots,\omega_n)$, we have
$\widetilde {\mathbb E}[Xf(\omega_{n+1})]=\widetilde {\mathbb E}[X\widetilde
{\mathbb E}[f(\omega_{n+1})|{\cal F}_n]]=\widetilde {\mathbb E}[X(\tilde p_nf(H)+\tilde q_nf(T))]$.
\end{remark}



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\end{document}
